We need answer math problem: "What is the largest k such that in any permutation of Z there must exist a monotone k-term arithmetic progression x1 < ... < xk?" Need interpret carefully. Permutation of Z: bijection f: Z -> Z? Or sequence (a_n)_{n∈Z} is a permutation of integers? Need "in any permutation of Z there must exist a monotone k-term arithmetic progression x1<... Z enumerated as (π(n))_{n∈Z}. We look for indices? "monotone k-term arithmetic progression x1 < ... < xk" likely terms x_i in the permutation (values?) such that x_i (indices? values?)? Let's infer from combinatorics: For permutations of N, existence of monotone arithmetic progressions? Could be: In any permutation of positive integers, there is a subsequence (not necessarily consecutive?) whose indices form an arithmetic progression and whose values are monotone? Wait "monotone k-term arithmetic progression x1<...0. So x_i are terms in the permutation? Or values? Need identify known problem. Question: "What is the largest k such that in any permutation of ℤ there must exist a monotone k-term arithmetic progression x1 <⋯ < xk?" Could refer to: Given a permutation p: Z -> Z (bijective ordering), consider the sequence p(0),p(1),...? A monotone k-term AP in a permutation means there exist i_1<...Z; "monotone k-term arithmetic progression x1<...Z. A monotone k-AP in π is positions a, a+d, ..., a+(k-1)d such that π(a), π(a+d), ..., π(a+(k-1)d) are monotone (increasing or decreasing). Or maybe values x_i form AP and their positions monotone? The notation x1<...Z, is there an arithmetic progression a, a+d,...,a+(k-1)d that appears in increasing order in the permutation (π^{-1}(a) < ...)? Or maybe just as a set? Since values can be arranged, need avoid long monotone AP. This resembles "van der Waerden for permutations" or "Erdos-Szekeres for arithmetic progressions". Need largest k maybe 3? 4? We need determine. Let's search memory internally. There is known problem: "What is the largest n such that every permutation of the integers contains an increasing arithmetic progression of length n?" Answer maybe 3? Or 4? There's a result: Every permutation of [N] contains an increasing arithmetic progression of length 3? Is that true? For permutations of [n], longest increasing AP? There is concept "increasing arithmetic progressions in permutations". Avoiding length k. What is maximum length forced in any permutation of integers? Might be 3 because can avoid 4? Need investigate. Alternative: If permutation of Z (two-sided) maybe can avoid monotone 4-term AP but any 3? Let's test. Interpretation A: A monotone k-AP in permutation = sequence of positions i_1 < ... < i_k with positions themselves in AP? Then need find largest k such that for any bijection π:Z->Z, there exist indices a,a+d,... (AP positions) with values π(a+jd) monotone (all increasing or decreasing). But any sequence of distinct integers of infinite length has infinite monotone subsequence, but subsequence indices not constrained to AP. Constrained to AP indices. Can we construct a permutation of Z such that for every AP of positions, values are neither monotone? Maybe finite? For length 2 always monotone? Any two numbers are monotone (x1). So k≥2. For k=3? Need every 3-term AP in indices yields values monotone (increasing or decreasing). Is that forced? No: permutation 1,3,2? For indices 0,1,2: 1,3,2 not monotone. But over bi-infinite maybe can avoid? Need largest forced maybe 2. But problem likely nontrivial so maybe answer 2? But question asks "largest k" maybe finite. Need determine. Interpretation B: Values x_1<...0? Impossible because if a and b appear, one before the other. For increasing AP length 2, we need two values a -i > -j => π(i)>π(j), no increasing pair. So no monotone increasing 2-term. If monotone allows increasing or decreasing, then decreasing pair exists. But notation x1<... -x2, so pos x1 after x2. Thus no increasing pair. So if order in permutation matters, largest k could be 1? But question likely assumes monotone in permutation order? Maybe "monotone" means indices? Let's be precise. The phrase "in any permutation of ℤ there must exist a monotone k-term arithmetic progression x1 <⋯ < xk" Maybe they define "monotone" as the terms of the permutation x_i (i is position?) satisfy x1<...Z. A monotone arithmetic progression of length k in π is a sequence x_1,...,x_k such that x_i = π(a + (i-1)d) and x_1 < ... < x_k. The x_i are the values in AP positions. Then x_i are terms of the permutation at AP indices; monotone values. They denote x_1<...b>c. Nonmonotone if b is middle? Wait a,b,c distinct. Monotone if b is not between a and c? Let's check: increasing: ab>c (b middle). So any three distinct numbers where the middle position value lies between the endpoint values is monotone? If ab>c, b between endpoints. If a nonmonotone. So nonmonotone iff the middle term is extremum (greater than both or less than both). So condition: For every 3-term AP of indices i,i+d,i+2d, the middle value π(i+d) must be either greater than both endpoints or less than both endpoints. This is like a sequence with all 3-term APs "peaks" or "valleys" at middle. Is that possible infinite? Maybe not? Let's check local d=1: for every consecutive triple, middle is local extremum. Thus sequence alternates high-low-high-low... i.e., no three consecutive monotone. Easy: e.g., odd positions high, even low. But for d=2: every triple with step 2 (positions i,i+2,i+4) middle position i+2 must be extremum relative to i,i+4. If parity pattern, positions same parity; can arrange maybe. Need all distances. This condition might be impossible beyond finite length? Let's see. For d=1, π(n) alternates local maxima/minima: sign π(n+1)-π(n) alternates. So sequence is like zigzag. For d=2, sign π(n+2)-π(n) ? Since π(n+2) must be extremum relative to π(n),π(n+4). This imposes pattern on same parity subsequences: each term in even subsequence at index n must be local extremum relative to neighbors two apart. Thus each parity subsequence also zigzags. In fact for every d, subsequence along AP with step d must have every triple nonmonotone, i.e., each interior term extremum relative to neighbors distance d. That means for each d, sequence (π(n+id)) has no monotone triple. Is there a bi-infinite sequence of distinct integers such that for every d, its d-spaced subsequence is zigzag? Maybe possible using ordering by 2-adic valuations? Let's explore. This resembles "3-free sequences" or "no 3-term monotone AP". Could be related to "Stanley sequence"? For any permutation of N, there must be a 3-term monotone AP? Maybe yes by Ramsey? Let's attempt prove: In any infinite sequence of distinct reals, for some d, triple (n, n+d, n+2d) monotone? We can choose any 3 equally spaced positions. Can a sequence avoid all? Let's attempt finite construction for length N? For N=3, nonmonotone possible. N=4 with d=1 only triples 1,2,3 and 2,3,4 nonmonotone: need signs alternate; possible: low high low high. N=5 with d=1 and d=2: sequence L,H,L,H,L? d=1 nonmonotone if middle extremum: pattern must alternate: positions 1,2,3: 2 extremum; 2,3,4: 3 extremum; 3,4,5: 4 extremum. So signs alternate: maybe values: pos2 high, pos3 low, pos4 high. d=2 triple 1,3,5: pos3 must extremum relative pos1,pos5. pos3 is low, so need pos1,pos5 > pos3. possible. Sequence low? Example pos1=2, pos2=5, pos3=0, pos4=4, pos5=3. d=1: 2,5,0 non; 5,0,4 non? endpoints 5,4, middle 0 less both non; 0,4,3 non? middle 4 > both endpoints? 4>0,3 yes. d=2: 2,0,3 middle 0 < both non. So length 5 possible. N=6? Need maybe possible. Could construct using binary reflected Gray? Let's formulate. Condition for no monotone 3-AP: For all n,d: π(n+d) > max(π(n),π(n+2d)) or < min(...). Equivalently, the sequence is a "3-free" or "alternating" along every AP. Maybe this is equivalent to π^{-1} order being a "3-free" linear order? Let's invert: A permutation is a linear ordering of Z. A monotone 3-AP in positions? Hmm. If no monotone 3-AP, for any three equally spaced positions arank b>rank c; in both, rank of middle is between ranks of endpoints. Nonmonotone: rank of middle is not between endpoints (it is min or max). Thus condition: For all APs of indices, the middle index's rank is an extreme among the three. So in the linear order of Z induced by the permutation (value order), no 3-term AP has its middle point between the endpoints in the linear order. This is exactly a "3-free" ordering: no 3-term arithmetic progression is ordered as x π(j) > π(k). These are permutations with no monotone 3-AP. Do they exist? I recall "A sequence of distinct integers is called 3-free if no three terms form an arithmetic progression in the order of the sequence"? There are constructions using base 3. But here indices are AP, values monotone. There is known concept: "3-free sequence" (Stanley sequence) is subset of N with no 3-term AP. But here permutation of all N with no monotone 3-AP is equivalent to a "3-free" linear order? I think possible: order integers by 2-adic valuation? Let's test. If such permutation exists, largest forced k would be 2 (if monotone allows inc/dec) or if only increasing, 1. But if no 3-free permutation of Z, then largest k at least 3. Need know. Let's try construct no monotone 3-AP on N (or Z). Suppose define order by binary reflected Gray code? Need for every triple i, j, k with j=(i+k)/2, j's order extreme. In terms of a total order <_π on indices (i <_π k iff π(i)<π(k)), condition: For all i fails. Maybe order by bit-reversal? Let's brute mentally. Try p(i)=bit-reversal(i). For 0,1,2,3: p:0,2,1,3. Triple 0,1,2: values 0,2,1 nonmonotone? 0<1<2? Actually positions 0 val0,1 val2,2 val1: 0,2,1 non (middle max). 1,2,3: vals 2,1,3: middle min. Only d=1. For d? 0,? 0,? (0,? 2) covered; (0,? maybe (0,? 3 no midpoint). Works for 4? p: [0,2,1,3] has no monotone 3-AP? Check (0,1,2): 0,2,1 non; (1,2,3):2,1,3 non. Good. For 8, bit-reversal of 0..7: 0,4,2,6,1,5,3,7. Check d=1: sequence zigzag? 0<4>2<6>1<5>3<7 yes. d=2: subsequence even indices: positions 0,2,4,6 values 0,2,1,3: triples 0,2,1 (middle 2 max), 2,1,3 (middle1 min) ok. Odd indices: 4,6,5,7: 4<6>5<7 ok. d=3: triples (0,3,6): 0,6,3 middle 6 max; (1,4,7):4,1,7 middle1 min; d? (2,? 8? no). Seems bit-reversal may work for powers of 2! For indices 0..2^m-1, bit-reversal order? Need check all APs. There is known: Van der Waerden? Bit-reversal permutation avoids 3-term monotone AP? Let's analyze. Let p(i)= bit reversal (with m bits). Claim: For any 0≤i 4,6,5 non (middle max). i=2,k=6,j=4: p=2,1,3 non. i=0,k=4,j=2: 0,2,1 non. i=0,k=6,j=3: 0,6,5 non? 0<5<6? Sequence 0,6,5: middle 6 max. i=1,k=7,j=4: 4,1,7 non. i=2,k=4,j=3: 2,6,1 middle max. Seems. Can prove: For binary addition without carry? If j midpoint means i+k=2j, so in binary j = (i+k)/2. The highest bit where i and k differ? Since ia_j>a_k. It is known that for every n there exists a 3-free permutation, maybe by a recursive construction. For n=2^m, bit-reversal works. For arbitrary n, truncate? Maybe. For Z, can extend two-sided via e.g. interleaving negative and positive bit-reversal? Need monotone 3-AP across zero? Need handle. If 3-free permutations of N exist, then largest k = 2 (for monotone inc or dec). If only increasing AP (not decreasing), maybe bit-reversal has some decreasing? But problem writes x1<... 0,4,2,6,1,5,3,7. Check all 3-term APs (i,j,k): d=1: 0,1,2: 0,4,2 -> 0<2<4? middle 4 max non. 1,2,3:4,2,6 -> middle 2 min non. 2,3,4:2,6,1 -> 6 max non. 3,4,5:6,1,5 -> 1 min non. 4,5,6:1,5,3 -> 5 max non. 5,6,7:5,3,7 -> 3 min non. d=2: 0,2,4:0,2,1 -> 2 max non. 1,3,5:4,6,5 -> 6 max? endpoints 4,5; middle 6 > both non. 2,4,6:2,1,3 -> 1 min non. 3,5,7:6,5,7 -> 5 min? endpoints 6,7, middle5 6 max. 1,4,7:4,1,7 -> 1 min. No d>3 for length8. So no monotone 3-AP. So if monotone inc/dec, k=2 can avoid 3. But if monotone increasing only? p bit-reversal has some increasing pairs? Need length3 increasing AP positions? For d=1: no increasing triple? all nonmonotone, but some decreasing/increasing? For no monotone 3-AP includes no increasing triple. So bit-reversal avoids increasing 3 too. For length2 increasing: there are many: 0,4. So if only increasing 2, k≥2? But -id avoids increasing 2 if two-sided. However bit-reversal indexed 0..n-1 not bi-infinite. For Z, can avoid increasing 2 via -id. So answer depends. The problem asks "monotone k-term arithmetic progression x1 <⋯ < xk". Monotone plus x1<... indicates increasing, not "inc or dec". But in combinatorics, a "monotone arithmetic progression" often means either increasing or decreasing, but if they write x_1<...0"? Then monotone means increasing. In number theory, a "monotone arithmetic progression" of length k is just a k-term AP with positive common difference, i.e., x_1<...0 and indices n_1<... Z is a permutation of positions. Need find a k-term arithmetic progression in the domain (values) that is monotone (increasing) in the range q. This is exactly monotone k-AP in a permutation q (positions as values, AP indices). It's the same as interpretation A but with inverse: indices are values in AP, values are positions monotone. Since p arbitrary, q arbitrary. So same condition: for every permutation q of Z, is there an AP of indices (values) of length k whose q-values are monotone (increasing if x1<... positions? If q(v) increasing). For monotone inc/dec, symmetric. For only increasing, q=-id avoids? q(v)=-v gives decreasing for any AP v_1<...Z can be -id. If only increasing, answer 1, not interesting. Unless "monotone" allows either inc or dec, so -id doesn't avoid. Then 2 or 3. Maybe "x1 < ... < xk" just lists the AP terms in increasing value order; "monotone" in permutation might mean their positions are monotone (either increasing or decreasing)? They might write "monotone k-term arithmetic progression x_1<... positions is a permutation with an AP of indices increasing in q. Since q can be any permutation of Z (two-sided), if only increasing, q=-id avoids all length≥2. If domain N (one-sided), q: N -> Z? Wait p: N -> Z bijection, inverse q: Z -> N not a permutation of N but a bijection from Z to N; can q(v)=-v? No positions N nonnegative, cannot all decreasing. Could still avoid long increasing AP? Maybe. But problem says permutation of Z, so domain and codomain both Z? Usually "permutation of Z" means bijection of Z with itself. Then inverse also Z->Z; q=-id possible. Thus only increasing trivial. Therefore they must mean monotone (inc or dec), or indices in AP and values monotone with both directions allowed? But x1<......>q(v_k). This matches interpretation A (with q) and allows -id to have monotone decreasing AP arbitrary, so not trivial. The condition "x_1<...Z positions. The sequence q(n) (n value) is a permutation of positions (Z). A monotone 3-AP in values with monotone positions means for some a,a+d,a+2d, the three q-values are monotone (middle q(a+d) between endpoints? For monotone inc/dec, middle position is between endpoint positions). We want avoid: for all a,d, q(a+d) is not between q(a) and q(a+2d) (i.e., is min or max). Same as 3-free permutation of the domain Z. So can we have a bi-infinite sequence q:Z->Z with no 3-term AP where middle term is between endpoints? This is a "3-free permutation" of Z. Finite 3-free permutations of intervals exist. Does a bi-infinite one exist? Maybe yes via a limit or recursive construction on Z. Need construct. If a bi-infinite 3-free permutation exists, largest k=2. But maybe no due to compactness? Finite intervals of arbitrary length 3-free may not extend to Z because APs can cross boundaries. But if for every finite subset there is a partial permutation with property, by compactness (finite intersection) there is a total order/permutation of Z? Need careful: We need a bijection q:Z->Z. If for every finite set F⊂Z there is a permutation of Z (or partial) such that all 3-APs contained in some large interval are nonmonotone, compactness over discrete product Z^Z? The property is local (finite conditions). If for every N there is a permutation q_N of Z avoiding monotone 3-AP on all APs within [-N,N], then by compactness (Z discrete noncompact? Product Z^Z not compact unless use one-point compactification? But finite constraints and existence of permutations? Could use inverse limits? Maybe construct explicitly.) Can we extend bit-reversal to Z? Let's try. Z can be partitioned into intervals [0,2^m-1], [-(2^m-1),-1]? Need ensure APs crossing. Maybe use a recursive "3-free ordering" of the integers: Order by absolute value? Let's test. Define order (permutation q values? Actually sequence q(n) positions) by 2-adic valuation? We need for all APs, middle is extreme. There is a known total order on Z called "2-adic order" that has no 3-term AP monotone? Let's investigate. Let rank of integer n by (-1)^? v_2(n) maybe. For any AP a,a+d,a+2d, one of the terms has strictly smaller 2-adic valuation than the other two? Is that true? For nonzero integers, among x, x+d, x+2d? Wait AP terms x, x+d, x+2d. The middle x+d. Need middle have extreme valuation, not just one of three. For 3 terms a,a+d,a+2d, the middle is a+d. Does v_2(a+d) always differ? Let's test: a=0,d=1: 0,1,2. v2(1)=0, endpoints 0 (∞),2(1): middle smallest if 0∞? Not strict? If use 0 as ∞, middle min. a=1,d=1:1,2,3: middle v=1, endpoints 0,0: middle max. a=2,d=2:2,4,6: middle v=2, endpoints 1,1: max. a=2,d=1:2,3,4: middle v=0, endpoints 1,2: min. Is it always true that v_2(middle) is not equal to both? It could equal: a=1,d=2:1,3,5 all v=0. Then tie. Need tie-break recursively perhaps by odd part. For any AP x, x+d, x+2d, if d is odd, parities: endpoints same parity, middle opposite? If d odd, x and x+2d same parity, x+d opposite. Thus middle v2=0 if odd; endpoints may be odd too (v=0) tie, or even (v>0). If endpoints odd, all odd, tie. Need higher scale: divide by? This suggests an ordering by the 2-adic absolute value (valuation) plus odd part recursively can make middle extreme. In fact, the 2-adic metric: In any 3-term AP in Z, the middle point is closer to one endpoint? There is an ultrametric property: For x,y,z with y=(x+z)/2, one of distances |x-y|_2, |y-z|_2, |x-z|_2? Since y-x = z-y = d. In 2-adic metric, |x-y|_2=|z-y|_2=|d|_2. The distance |x-z|_2=|2d|_2 ≤ |d|_2, strict if v2(d) finite? If d even, |2d|_2 < |d|_2; if d odd, equal. In an ultrametric, if two sides equal and base smaller/equal, the middle? We want order by distance from 0 maybe? Not. Maybe use 2-adic valuation of the index in a "Farey" order. There is known "2-adic ordering" of integers: order by increasing 2-adic absolute value? It has property that no 3-term AP is monotone? Let's test on N. Define order by v2(n) ascending (0 before 1 before 2 ...), ties by something. Sequence of natural numbers by v2: odds (v0), then numbers v1 (2,6,10?), etc. Does this avoid monotone 3-AP? For AP 1,3,5 all v0; tie-break among odds maybe by odd value. If tie-break normal order, 1<3<5 monotone (bad). Need recursive: order odds by their odd part after mapping? Since odds form a copy of Z under n -> (n-1)/2? We can recursively order. This suggests a recursive construction: To order a set S such that midpoints extreme, choose an involution? For Z, partition into odd/even? For any AP: - If common difference d is odd: endpoints have same parity, middle opposite. Then if we order all numbers of one parity before the other? Let's see. Suppose all evens before all odds. For d odd: endpoints even, middle odd (if start even) or endpoints odd, middle even. If evens before odds: endpoints before middle => sequence positions: endpoints low, middle high; middle is max? For endpoints even: pos(end) < pos(middle), so in triple (end1, middle, end2), positions (low, high, low): middle max, nonmonotone. For endpoints odd: positions (high, low, high): middle min, nonmonotone. Good! So separating parity with evens before odds handles odd d. - If d is even: all three terms have same parity. Then divide by 2: (a/2, a/2+d/2, a/2+d) if a even, etc. The condition reduces to ordering within each parity class. Thus recursively order evens (as copy of Z) and odds (as copy of Z) using the same rule, but we also need a global order that puts one parity before the other? Wait if we put all evens before all odds, then within evens the order is a 3-free permutation of Z (scaled), and within odds likewise. For d even, triple lies in one parity, and positions are within that block; middle extreme by recursive property. For d odd, parity separation ensures middle extreme. This gives a bi-infinite 3-free permutation of Z! Let's construct explicitly: Define total order <_Q on Z: first all even integers (in the order induced by scaling and <_Q), then all odd integers (in order induced by scaling and <_Q). But to be a permutation (sequence) of Z, we need a linear order with all evens before all odds, and within evens an ordering of evens, within odds an ordering of odds. Since Z is bi-infinite, "all evens before all odds" in a linear order of type η? Wait a linear order on Z (the set of values) with all evens before odds and each parity bi-infinite. That's order type (Z, order within evens) + (Z, order within odds). But each parity with recursive order type? If base order is Z type? Let's define a total order (not a sequence indexed by Z?) Actually a permutation of Z is a bijection p:Z→Z, equivalently a linear order of type Z (order type of integers) on the set of values? Wait a sequence indexed by Z gives a linear order of type Z: ... < ... < ... < ... with each element finitely many predecessors? A permutation of Z as a bijection q:Z→Z: indices n∈Z ordered by usual Z; q(n) are values. The order of values induced by positions is a linear order of type Z (bi-infinite with no endpoints). It is not any linear order; it must have order type Z (order-isomorphic to usual Z), i.e., discrete, each element has immediate predecessor/successor, and both directions infinite. "All evens before all odds" with each parity block order type Z yields order type Z + Z, which has a cut with no immediate transition? Actually Z + Z is a linear order with a last even? No, Z has no last element, so the concatenation Z + Z has no immediate boundary but order type is η? Wait Z (order type ω*+ω) + (ω*+ω) = ω* + (ω+ω?) Let's compute: usual Z = ω* + ω (nonpositive then nonnegative). Concatenate two copies: (ω*+ω) + (ω*+ω) = ω* + (ω+ω*) + ω. The middle ω+ω* is dense? No, ω+ω* has no adjacency between the end of first ω and start of second ω*? First ω has no last, second ω* has no first, so there is a gap (a convex subset?) The order is not Z because there is a Dedekind cut with no adjacent elements and both sides nonempty. It is not discrete. A permutation indexed by Z must have immediate successor for each position. Thus cannot simply put a whole infinite block before another block in a bi-infinite sequence. Important! For finite intervals, we could place evens before odds within finite block; for Z, all evens before all odds is not order type Z. Need a permutation sequence indexed by Z, i.e., a discrete bi-infinite order. Recursive parity block not directly possible unless interleave blocks in a way preserving discrete order. Maybe use a "tree" of dyadic intervals: define order via 2-adic valuation but as a sequence indexed by Z? The natural 2-adic order on integers? Let's think. A permutation of Z is a discrete linear order of type Z. Can we color evens/odds alternately to handle odd d? For odd d, endpoints same parity, middle opposite. If in the sequence order, the middle parity class element is between endpoints? We need middle position extreme relative to endpoints, not necessarily block. If we can ensure that for any a of one parity and endpoints of opposite parity (a±d), the position of a is outside interval between positions of endpoints. That is hard globally. Maybe use a "alternating" sequence where parity alternates each step: positions n and n+1 have opposite parity. Then for d=1 (odd), middle position n+1 has parity opposite endpoints n,n+2 (same parity); but position of middle is between endpoints in the index order (since positions are consecutive), so in the sequence order, middle position is between endpoints! Wait we are considering q-values as positions? Let's fix. We want q: value n -> position q(n). The order of values by positions is a permutation sequence. For a 3-AP in values a,a+d,a+2d, we look at their positions q(a),q(a+d),q(a+2d). The "middle" in AP is a+d. For it to be nonmonotone, its position must be outside the interval between the endpoint positions. If we interleave parity by position, not value, not enough. In finite bit-reversal, parity of index and value? Let's see p index (value?) bit-reversal q? For q(n)=bit-reversal(n) on finite [0,8). For d odd, endpoints same parity, middle opposite. In q positions, is middle position outside endpoints? Example q(0)=0,q(1)=4,q(2)=2: middle position 4 outside [0,2]? yes > both. q(1)=4,q(2)=2,q(3)=6: middle 2 outside [4,6]? < both. It alternates. Parity separation by value? q positions of evens: 0,2,1,3? not all before odds (odds 4,6,5,7) actually evens before odds for first 8! In finite sequence of length 8, all even values positions 0-3, all odd values positions 4-7. That works for finite. For infinite, cannot all evens before odds if order type Z? But maybe we can use a bi-infinite sequence that is a limit of finite blocks where evens before odds at each scale, with a "middle" moving to infinity? For Z, maybe use a 2-adic tree traversal that is discrete and each parity not a convex block. Let's explore. Can there be a discrete bi-infinite order of Z with no monotone 3-AP? The finite recursive construction suggests using order type of a binary tree with no adjacent gap? Need a "2-adic" ordering of Z that is discrete of type Z. Maybe the order of integers by 2-adic valuation in a Cantor-like interleaving: odds and evens interwoven recursively, but still discrete. For example, order: ...,? Let's try define q(n) = some integer based on binary representation (two-sided) such that q(a+d) extreme. Maybe use the "Thue-Morse" ordering. The bit-reversal finite permutation can be extended to Z via the 2-adic integers? Consider q(n) = bit-reversal of the 2-adic representation? For n∈Z, write in binary with sign? The order by 2-adic valuation might yield a discrete order? Let's investigate. There is a known total order on Z: the 2-adic order (or "odds before evens" recursively) but as a linear order of type η? Actually the order type of Z_2? The 2-adic integers with lex order? Hmm. But a permutation of Z must be a bijection from Z to Z, not just any linear order. However any countable dense order without endpoints can be interleaved? Wait order type of a sequence indexed by Z is exactly Z: discrete, each element has immediate predecessor and successor, and no endpoints. If we define a total order on values, it must be order-isomorphic to Z. The finite recursive "evens before odds" gives order type finite; infinite recursive gives maybe order type of the 2-adic integers? Not Z. Could we still index a non-discrete order by Z? No, the position order is always discrete. So if the necessary recursive condition forces a non-discrete order, no bi-infinite 3-free permutation. But maybe there is another construction that is discrete. Let's analyze impossibility of 3-free permutation of Z. Suppose q:Z->Z is a permutation (sequence indexed by Z). For every 3-AP in domain, q(middle) extreme. In particular for d=1: for every n, q(n+1) is either > q(n),q(n+2) or < q(n),q(n+2). Thus signs s_n = sign(q(n+1)-q(n)) must alternate: if q(n+1) is local max, s_n=+, s_{n+1}=-; if local min, s_n=-,s_{n+1}=+. So s_{n+1} = - s_n for all n. Therefore the sequence q(n) is a strict zigzag: signs alternate. Thus q(n) for even n is either all local maxima and odd local minima, or vice versa. So q(2m) are high relative to neighbors, q(2m+1) low relative to neighbors (or opposite). This imposes that even subsequence and odd subsequence each? For d=2: q(n+2) extreme relative to q(n),q(n+4). This means within even positions (as function m->q(2m+1?) wait middle n+2 same parity as endpoints) the sequence over same parity also zigzags. Specifically for even subsequence E_m=q(2m), for any m, E_{m+1} extreme relative to E_m,E_{m+2}. Thus signs within E alternate. Similarly odd subsequence. By induction, for every d, the subsequence q(n+kd) is zigzag: signs along step d alternate. This is a strong condition. Can a bi-infinite sequence of distinct integers satisfy that for every d, its d-th decimation is alternating (no monotone triple)? Let's call it "strongly zigzag". Does such a sequence exist? This is equivalent to a permutation of Z with no monotone 3-AP. Let's attempt construct or prove impossible. Let q:Z->Z. Define for each n,d the sign σ(n,d)=sign(q(n+d)-q(n)). The condition that q(n+d) is extreme in triple (n,n+d,n+2d) means σ(n,d) = - σ(n+d,d). Because if q(n+d)>q(n) and >q(n+2d), then q(n+d)-q(n)>0, q(n+2d)-q(n+d)<0, so σ(n,d)=+, σ(n+d,d)=-. If middle min, σ(n,d)=-, σ(n+d,d)=+. Thus condition: σ(n+d,d) = - σ(n,d) for all n,d. (1) This is exactly: For each fixed d, signs along n alternate. So σ(n+d,d) = (-1)? We can derive σ(n+2d,d)=σ(n,d). Therefore for fixed d, sign of difference over step d has period 2 in n: σ(n+2d,d)=σ(n,d). Also σ(n+d,d)=-σ(n,d). In particular, for any n, σ(n,d) depends only on parity of floor(n/d)? But need consistency across d. Can we assign σ(n,d) = (-1)^{?} satisfying additive relations? Since q is a function, differences must be consistent: For any m,n, q(n)-q(m) sign may be determined by path? But σ only for specific equal steps. Let's use d=1: σ(n,1)=(-1)^{n} times a global sign (alternating). So q(n+1)-q(n) alternates sign. Thus q has local extrema. Let's assume q(0)q(2)... or opposite. WLOG maybe q(1)>q(0) and q(1)>q(2); then q(n+1) = -q(n) for n even? More precisely σ(0,1)=+ if q1>q0; σ(1,1)=-; so q(n+1)-q(n) = (-1)^n c? With c=+ for even n. We can reverse value order to set c=+. Now consider d=2: σ(n,2) alternates in n with period 4. It is sign(q(n+2)-q(n)). For n=0: q(2) vs q(0). Since q1 is max of q0,q2, q2q0 or ? If r positive increasing? Example finite bit-reversal not simply. Suppose q(n)=(-1)^n A(n) where A(n)>0? For d odd, q(n+d) has opposite sign to q(n). Then σ(n,d)=sign(opposite sign - same sign) always negative if A positive? If q(n)=+A_n for even, -A_n for odd. For d odd, q(n) even positive, q(n+d) odd negative => σ=-; q(n+d) to n+2d odd negative to even positive => σ=+. So alternates automatically for all odd d! Good. For d even, q(n) and q(n+d) same sign; sign depends on A. Thus we need the even subsequence A(even) and odd subsequence A(odd) each satisfy no monotone 3-AP (since for d even, divide by 2). This suggests recursive construction: q(n) = (-1)^n * Q( floor(n/2) ) maybe? For even n=2m: q=+Q(m). For odd n=2m+1: q=-Q(m). If Q:Z->? is a 3-free sequence? Let's check. Define q(2m) = Q(m), q(2m+1) = -Q(m). But codomain must be Z permutation: if Q is permutation of positive? We need all integers once. This maps pairs to ±Q(m). If Q(m) ranges over positive? Then q ranges over all nonzero with duplicates? Need include 0. Could map Q:Z->Z nonzero? Let's design: For each m, assign a pair of opposite values: +a_m and -a_m. If {a_m} is a set of positive integers (all positive), q is a permutation of Z\{0}. Add 0 somewhere? But 0 would break signs. Maybe use a permutation of nonzero integers paired by ±. For any m, choose a distinct positive integer a_m; q(2m)=a_m, q(2m+1)=-a_m. This is a bijection from Z to Z\{0}; not include 0. Could shift? Maybe use a permutation of Z where no zero? Need include 0; can put 0 at one position and adjust? Or use pairs (a_m, b_m) not opposite but all positive vs all negative? For odd d, need endpoint signs opposite and magnitudes not matter: even values all positive, odd all negative works for odd d. Then values used: positive integers at even positions, negative at odd positions. Zero? Could assign 0 to some position, but then sign argument for APs involving 0 may fail. Maybe avoid 0 by using positive and negative including 0 as one of negatives? If one parity contains 0 and positives, other negatives? For odd d, middle value must be extreme relative endpoints. If endpoints one positive one negative? Wait AP with d odd: endpoints same parity. If even positions positive, odd negative. For AP with endpoints even (positive), middle odd (negative): middle is less than both endpoints -> min, good. For endpoints odd (negative), middle even (positive): middle max, good. This holds regardless of magnitudes. If one of values is 0, still if endpoints positive and middle negative/0? If middle 0 and endpoints positive, 0 min if positive; if endpoints negative and middle 0, 0 max. If endpoints include 0? endpoints same parity; if one parity contains 0 and positive, endpoints may be 0 and positive, middle negative: middle < both? If endpoint 0, negative <0 good. If other parity contains negative only. So we can put all nonnegative in even positions, all negative in odd positions. Then odd d condition holds. For even d, endpoints and middle same parity; within nonnegative or negative, need recursive property. This maps even positions to nonnegative integers, odd to negative integers. But even positions are bi-infinite (Z), nonnegative integers one-sided (N). Cannot biject Z positions to N. Need each parity class (infinite in both directions) map to a set of order type Z? For a permutation of Z, values in even positions can be any subset of Z of cardinality continuum? Countably infinite. It can be two-sided (e.g., all evens) to allow recursive scaling. The sign trick requires even values all > odd values (or <), but the even subset must be infinite in both "directions"? It just needs to be a copy of Z for recursive 3-free property, but also lie entirely on one side of all odd values to handle odd d. Can we partition Z into two subsets A,B such that every element of A > every element of B, and both A and B are in bijection with Z? No, because no infinite initial segment? If A all > B, then A has a lower bound? In Z, if every a∈A > b∈B, then B is bounded above by min A? But min A may not exist; e.g., A={1,2,3,...}, B={-1,-2,...}; A one-sided, B one-sided. Not both order type Z. If A=positive, B=negative plus 0, A one-sided. Cannot have both copies of Z separated by a cut because Z has no infinite gap; a cut with both sides infinite in Z is between n and n+1, one side one-sided? Actually in Z, a convex initial segment either finite/cofinite? If A is upper set and B lower set, if both infinite, there is a cut at infinity? No, in Z any partition into lower and upper sets with all A>B has a threshold: B⊂(-∞,t], A⊂[t+1,∞); one or both may be infinite but each one-sided, not both two-sided. So cannot recursively scale with same sign separation in Z. For finite intervals, can split lower half/upper half. For Z, no middle split into two Z copies with all one side lower. This hints no bi-infinite 3-free permutation, maybe largest k≥3. But wait finite bit-reversal used evens (values) before odds (positions) but value set [0,8) split into lower/upper halves? Actually q values positions 0..7; domain values 0..7. Even domain values (0,2,4,6) got lower positions 0..3; odd domain values got higher positions 4..7. So in the order of positions, all even values come before all odd values. For finite domain, order type finite, possible. For infinite Z, order type Z cannot have a cut with both sides infinite without adjacency? It can have a cut but not both sides infinite? In order type Z, any proper initial segment is either finite or cofinite? Let's check: In usual Z order, an initial segment {x: xZ distinct. For any n, d, e, signs may satisfy multiplicative relations if steps combine. For example, consider d and d+? We can use d=1 and d=2. Let's set local pattern. WLOG q(0)q(2)... (alternating starting with max at odd? Let's fix q1 max between 0,2; q2 min between 1,3; etc.) So for all n even: q(n+1)>q(n), q(n+1)>q(n+2)? Wait if pattern +,-,+,-: σ(n,1)=+ for n even, - for n odd. Thus for n even: q(n+1)>q(n) and q(n+1)>q(n+2)? σ(n+1,1)=- means q(n+2)q2, so q2 local min. So even positions are local minima, odd local maxima. Now for d=2, σ(n,2)=sign(q(n+2)-q(n)). Since even positions are local minima relative to adjacent odd, but q(n+2) vs q(n) unknown. Condition: σ(n+2,2)=-σ(n,2). Thus for even n, the signs between consecutive even terms alternate: if q(2)>q(0), then q(4)q(4), etc. So even subsequence itself zigzags. Similarly odd. Can we assign values recursively to satisfy all d? This is like finding a linear order of Z such that each d-decimation is alternating. This may be possible using a "universal" ordering from a 2-adic tree but with discrete order? Let's attempt construct explicitly via q(n)=? Maybe q(n) = (-1)^n * 2^{v_2(n)}? Not permutation. Let's try finite infinite construction using "interleaving" finite 3-free permutations. For each N, we have 3-free permutation of interval [-N,N] maybe. Can we take a diagonal limit to get a permutation of Z? Since the property is local, if for each finite interval we can extend to a permutation of that interval (not necessarily Z) with no monotone 3-AP inside, then by compactness of the space of linear orders? But the order type must be Z (discrete). Finite approximations with boundaries may not converge to a discrete bi-infinite order; but perhaps we can construct an infinite sequence directly by interleaving finite blocks. Try to build a bi-infinite zigzag sequence recursively by blocks. For length 8 bit-reversal: 0,4,2,6,1,5,3,7. This can be seen as: lower half (0,2,4,6) mapped to first half by bit-reversal? Actually q(0..3)=0,4,2,6? Wait q on domain 0..7: positions: n:0 1 2 3 4 5 6 7 q:0 4 2 6 1 5 3 7 Even n (0,2,4,6) positions 0,2,1,3? Let's list q(0)=0,q(2)=2,q(4)=1,q(6)=3. This is not bit-reversal of half? If scale even n=2m, m=0..3, q(2m)=? 0,2,1,3 which is bit-reversal of m? For 4: bit-rev m:0,2,1,3 yes. Odd n=2m+1: q=4,6,5,7 = 4 + bit-rev(m). So finite construction: q_{m+1}(2m)=q_m(m), q_{m+1}(2m+1)=2^m+q_m(m), with q_1(0)=0. This is bit-reversal. It splits domain into evens/odds and positions into lower/upper halves. For infinite, this recursion would produce a map from Cantor tree to positions in [0,1] maybe, not a permutation of Z. Can adapt to Z by using a bi-infinite binary tree? Domain Z can be partitioned recursively into even/odd, but positions must be partitioned into two sets that interleave, not halves, to maintain order type Z. For odd d, we need if n and n+2d same parity, middle position outside endpoints. Perhaps if positions of evens and odds alternate? Let's test a simple candidate: q(n) = n? Then APs monotone (bad). q(n)=(-1)^n n? Not permutation. q(n) = (-1)^n floor? Need bijection. Maybe use q(n) = (-1)^n * (some rearrangement of positive integers). For odd d, if even values positive, odd negative, middle extreme. But as noted, even positions (domain n even) is a two-sided index set, but values positive one-sided. That's okay: q is a permutation of Z if even positions take all positive integers and odd positions all nonpositive? But there are countably infinite even positions (Z) and positive integers (N); bijection possible (Z to N) yes, e.g., enumerate positives over even indices in a zigzag. Odd positions take nonpositives (N*) bijective. Then q is a permutation of Z (even positions positive, odd positions nonpositive). Does this avoid odd d? For AP with d odd: endpoints same parity. If endpoints are even -> values positive; middle odd -> nonpositive (≤0). If nonpositive could be 0 and positives >0, middle ≤0 < both positives, min, good. If endpoints odd -> values ≤0; middle even positive, max, good. Thus all odd d satisfied regardless of ordering within parity. For even d: endpoints and middle same parity. Within even positions, the values are positive integers arranged by q(2m). Need no monotone 3-AP among APs in domain with even d. If a,a+d,a+2d all even, then divide by 2: m,m+e,m+2e where e=d/2. The positions q(2m) = some permutation P of positive integers (N). We need for all m,e, P(m+e) extreme relative P(m),P(m+2e). That is a 3-free permutation of N (one-sided) with values positive, avoiding monotone 3-AP (inc/dec). Similarly odd parity with nonpositive values: q(2m+1) = -P'(m) maybe; monotone condition for negative values reverses? If values all negative, increasing/decreasing reversed? But extreme property is invariant under negation? If sequence r(m) has no monotone triple, then -r also no monotone triple. So need a 3-free permutation of N for each parity. Thus if there exists a 3-free permutation of N (one-sided) of the positive integers, we can construct a 3-free permutation of Z by putting one parity on positives and the other on nonpositives! This seems promising. Earlier I thought Z cannot split into two Z copies by value separation, but here domain parity classes are indexed by N? Wait domain parity classes: even integers n=2m with m∈Z (bi-infinite). If we map even positions to positive integers, the index m ranges over Z, not N. The even subsequence over m∈Z must be a 3-free permutation of positive integers? But a sequence indexed by Z of positive integers cannot be a permutation of positive integers (which is one-sided) if it's bi-infinite? It can be a bijection from Z to N: yes, e.g., enumerate all naturals in a bi-infinite sequence. But does a "3-free permutation of N" usually indexed by N? We need a bi-infinite sequence P_m (m∈Z) that is a bijection onto N and has no monotone 3-AP for all m,e∈Z. Is that possible? The condition is local; one-sided vs two-sided may differ. We can perhaps take a one-sided 3-free permutation of N and extend to bi-infinite by padding? But N has no negative indices; APs can go both directions. A bijection Z->N has no endpoints; can it avoid monotone 3-AP? Maybe easier/harder. But we can choose even positions m∈Z to enumerate positive integers in a 3-free bi-infinite way if exists. Alternatively partition domain Z into two sets each bi-infinite; map one to positive N, other to nonpositive. Need 3-free sequences indexed by Z onto N and onto nonpositive. Does such exist? We can perhaps use the finite bit-reversal limit to get a sequence indexed by Z? Hmm. Maybe simpler: Use a 3-free permutation of N (indexed by N) for positive values and place it on even positions n≥0? But even positions include negative n; APs with even d can cross from negative to positive indices. We need full bi-infinite. Could choose a bijection from Z to N that is a "3-free permutation" of N in the bi-infinite sense. Is there a known existence? Let's attempt construct via the same sign recursion: For a sequence P_m (m∈Z) of positive integers, no monotone 3-AP. For d=1 in m, P must zigzag. A bi-infinite sequence of positive integers with no monotone 3-AP? Is that possible? Since positive integers are unbounded above but bounded below. A bi-infinite zigzag sequence of distinct positive integers: local minima and maxima. Can it extend infinitely in both directions? Yes, e.g., ... 5,1,6,2,7,3,8,4,...? Need all d. But maybe possible. However if values are bounded below, a bi-infinite sequence with local minima alternately maybe can exist by using arbitrarily small? Minimum 1 only once; local minima can be increasing? Let's see. A bi-infinite strictly zigzag sequence of positive integers: signs alternate. It has infinitely many local minima and maxima going to -∞ in index. Since values positive, local minima cannot go to 0 limit? They are integers ≥1. Infinitely many distinct positive local minima; possible (1,2,3,...). Maxima also distinct. No immediate contradiction. For d=1 only, sequence ... 2,5,1,4,6,3,7,... can zigzag. Can we construct bi-infinite 3-free P of N? Maybe use a permutation of N with no monotone 3-AP that is extendable. There are known "3-free permutations" of N for one-sided; can we make it two-sided by taking a bi-infinite sequence whose restriction to N is 3-free? APs with negative indices might be handled by a symmetry. Maybe there is a direct construction for Z using values split positive/nonpositive and each parity uses a "3-free" bi-infinite sequence of N. But if such P exists, then q exists. If not, maybe no. Let's step back. Maybe the answer is not 2 but 3 or 4. Need identify known theorem. Let's search memory more. Problem statement: "What is the largest k such that in any permutation of ℤ there must exist a monotone k-term arithmetic progression x1 <⋯ < xk?" This sounds like a question about "monotone arithmetic progressions in permutations of the integers". There is a known result by? "Every permutation of the positive integers contains a monotone 3-term arithmetic progression, but not necessarily 4"? Or "the largest is 3"? Let's test with finite 3-free permutations of [n]. If finite 3-free permutations exist for arbitrarily large n, then for any finite initial segment of a permutation of N, there is no forced 3 within that segment. But in an infinite permutation of N, maybe global compactness yields a 3-free permutation of N? If finite 3-free permutations of [n] exist for all n, can we get a permutation of N with no monotone 3-AP? Need ensure APs not crossing finite boundaries. Compactness over permutations of N? The set of all permutations of N is not compact in product topology? It is a closed subset of N^N with finite intersection? The space of injective sequences? For finite constraints, if for each n there is a permutation of N avoiding all monotone 3-APs with positions ≤n, then by König's lemma (compactness of product of discrete finite? N not compact) we can get an injective sequence? Need range constraints: values may escape to infinity. But a permutation of N requires all values used. Local avoidance plus surjectivity? Surjectivity is global (for each value v, some position). We can enforce value v appears in first n positions? Finite conditions may not be finitely satisfiable with permutations. Could use a compactness argument on linear orders: The set of linear extensions of a finite partial order? Hmm. If for every finite set of indices there is a partial permutation of those indices avoiding internal APs, can extend to full permutation by arbitrary placement of other values? But other values might create APs with the finite set. Need all APs. Could use a greedy construction? Maybe. Known: There exist "3-free permutations" of the positive integers? Let's recall. A permutation of [n] with no 3-term monotone AP is called a "3-free permutation". I think it is known that 3-free permutations exist for all n. For example, the "bit-reversal" permutation for n=2^m, and for general n by taking initial segment? Does initial segment of bit-reversal avoid? Not necessarily APs near boundary? But can construct recursively for all n. If they exist for all finite n, does it imply a 3-free permutation of N? I think yes via a compactness/limit argument: Consider all permutations of N as points in the product space of linear orders? There is a result: If every finite subset of N has a 3-free ordering, then there is a 3-free ordering of N? But the ordering must be of type ω, not just finite. Finite orderings can be extended by putting new largest element in appropriate place? Maybe. Let's try construct a 3-free permutation of N explicitly. Recursion: For n=2^m, bit-reversal. For infinite N, we could take the limit of bit-reversal as m→∞: This yields a linear order on the finite binary strings, i.e., on the 2-adic integers? The order type is not ω. But maybe we can enumerate it in a discrete ω order? Wait a permutation of N is just a sequence; the order of values by positions is type ω. If we take the limit of finite bit-reversal orders, we get an order on N that is not type ω (probably dense? or has jumps). But we only need the sequence positions; if the limit order is not type ω, cannot be a permutation of N? Actually any countable linear order can be listed in a sequence, but the sequence index order is ω; the induced order of values by sequence is the list order, which is type ω by definition. The finite limit order would be the order of values as they appear in the sequence; if it's not type ω, it cannot be the list order of a permutation. But the finite bit-reversal sequences have order type finite; their limit as sequence maybe a permutation of N? Let's see: q_m: [0,2^m) -> [0,2^m) bit-reversal. For m increasing, does q_m(i) converge for each i? No, values change unboundedly. But maybe define q(i) via binary expansion: q(i) = limit of bit-reversal? For i with finite binary, as m grows, q_m(i) = 2^m - bitrev_m(i)? Actually for fixed i, q_m(i) tends to infinity (since higher bits). Not a permutation of N. Maybe define q(i) using 2-adic representation reversed to get a 2-adic integer, not natural. Let's search memory: "There is no permutation of the natural numbers which is 3-free" maybe due to a theorem by D. E. Knuth? Let's test with sign equations. Suppose q:N->N (one-sided) no monotone 3-AP. For d=1, signs alternate. So q has local extrema. Since domain starts at 0, pattern either q0q2<... or q0>q1... . Consider local minima. If pattern q0q2..., minima at even positions (including 0? q0 endpoint). There are infinitely many local minima q(2n). They are distinct positive integers. The minimum of all these minima is some q(2n0). For nN 3-free. Choose n such that q(n) is the global minimum (value 1). Since no monotone 3-AP, for any d, if n is the middle of a 3-AP (n-d, n, n+d within domain), q(n) must be extreme. Since 1 is global minimum, it is automatically extreme (min) for any triple where it is middle. So no contradiction. If n is endpoint of a 3-AP: (n, n+d, n+2d) with q(n)=1. For monotonicity avoidance, the middle q(n+d) must be extreme relative 1 and q(n+2d). Since 1 is min, if q(n+d) is max? It could be >q(n+2d) (middle max) or q(n+2d). Thus for every d such that n,n+d,n+2d in domain, q(n+d) > q(n+2d). That means the tail after n is strictly decreasing? For all d≥1, q(n+d) > q(n+2d). This implies q(n+1)>q(n+2)>q(n+3)>... (taking d=1 gives q(n+1)>q(n+2); d=2 gives q(n+2)>q(n+3)? Wait d=2: q(n+2) > q(n+4); not adjacent. But combine? Let's check: For endpoint at n, triple (n, n+d, n+2d) with d=1: q(n+1) > q(n+2). d=2: q(n+2) > q(n+4). d=3: q(n+3)>q(n+5). Does this force entire tail decreasing? d=1 gives adjacent pair (n+1,n+2) decreasing; but for pair (n+2,n+3), use triple (n+1? no endpoint 1?). If 1 at n, only triples with n as endpoint. But monotone triples with middle n? OK. Need other endpoint? Maybe choose global minimum and consider it as endpoint for APs starting at n; gives constraints. For d=1: q(n+1)>q(n+2). For d=2: q(n+2)>q(n+4). Not q(n+3). But d? To compare q(n+2),q(n+3), consider triple (n+1,n+2,n+3) with middle n+2. It must be extreme. q(n+1)>q(n+2) from d=1. If q(n+2) were >q(n+3), then q(n+2) max, OK; if q(n+2)q(n+2)b and b2<3 nonmonotone (middle min). So both a>b>c (decreasing) and a>bb>c or ab, monotone only if b>c (decreasing). Nonmonotone if bb, to avoid monotone decreasing, need bq(n+2), no relation q(n+2),q(n+3). d=1 local zigzag? If q(n)=1 and d=1 signs? Let's set pattern. If q(n) is endpoint not necessarily local? For triple (n,n+1,n+2), to avoid monotone increasing (since q(n)=1 min), need q(n+1)>q(n+2) (middle max). For triple (n+1,n+2,n+3), with q(n+1)>q(n+2), to avoid decreasing, need q(n+2)q(n+2)q(n+4). Thus signs alternate: +? q(n+1)>q(n+2) (-), q(n+2)q(n+4) (-). So tail after n is zigzag starting with down. This is consistent with global d=1 alternation; q(n) as min endpoint forces q(n+1)>q(n+2). If pattern before n maybe. No contradiction yet. Could use global minimum 1 as endpoint to force for every d: q(n+d) > q(n+2d). In particular, for d=1: q(n+1)>q(n+2); d=2: q(n+2)>q(n+4); d=3: q(n+3)>q(n+5); etc. Combined with zigzag, maybe forces some value to be infinite? Let's examine. Let t_m = q(n+m) for m≥1. Constraints: t_1>t_2, t_2>t_4, t_3>t_5, t_4>t_6, etc. And zigzag: t_1>t_2t_4... (assuming). So t_2 < t_3, t_2 < t_1, t_2 < t_4? constraint t2>t4, so t2>t4. t2 is min among t1,t3,t4? t4t4t6. So t6t6>t7? If t6 local min, t6>t7? Wait pattern: t1> t2< t3> t4< t5> t6< t7>... Actually signs: t1>t2 (-), t2t4 (-), t4t6 (-), t6 t_{2d}. For even d=2: t2>t4 (min > next min? possible); d=4: t4>t6; d=6: t6>t8; thus even subsequence t_{2m} is strictly decreasing! Since t2>t4>t6>... positive integers, cannot be infinite. Aha! For N (one-sided), global minimum at n gives infinite decreasing sequence of positive integers, contradiction. Therefore no 3-free permutation of N (one-sided) with positive values! Nice. For Z, values are unbounded below, so a decreasing infinite sequence of integers is possible. Thus bi-infinite Z may allow 3-free permutations. Good. This suggests answer for Z might be 2 if we can construct, because the obstruction for N (well-ordering) removed. The problem specifically says ℤ, not ℕ, likely because 3-free permutations exist on ℤ but not on ℕ? Or because answer differs? Let's explore. Can we construct a bi-infinite 3-free permutation of Z using the sign recursion and unbounded both directions? Maybe yes. The finite bit-reversal can be shifted to bi-infinite? Let's try using values split positive/nonpositive and parity classes indexed by Z with sequences of N and -N-1 that are 3-free bi-infinite. Need construct a 3-free bi-infinite sequence P_m (m∈Z) of positive integers (or N) and Q_m of nonpositive. Maybe the same well-ordering obstruction disappears because P_m is bi-infinite but values positive; can we have a bi-infinite sequence of positive integers with no monotone 3-AP? Let's test using global minimum 1. In a bi-infinite sequence P_m of positive integers, value 1 occurs at some index n. Since indices extend both directions, consider triples with n as endpoint in both directions: (n-d, n, n+d) with n as right endpoint? Need avoid monotone. For triple (n-d, n, n+d) where n is middle? Actually if n is endpoint: AP (n-d, n, n+d) has middle n, not endpoint. For n as endpoint, APs (n, n+d, n+2d) to right and (n-2d, n-d, n) to left. Since q(n)=1 global min, for right endpoint constraints: for all d>0, P(n+d) > P(n+2d). This gives even subsequence to right decreasing positive: P(n+2)>P(n+4)>... contradiction? Wait in bi-infinite, right side still infinite; if P is a permutation of positive integers (bijection Z->N), it is unbounded above, but positive integers well-ordered, so an infinite strictly decreasing sequence of positive integers is impossible. Thus no bi-infinite sequence of positive integers with no monotone 3-AP! Because global minimum 1 as endpoint to the right gives infinite decreasing even subsequence. Let's verify: If P(n)=1, for every d≥1, triple (n, n+d, n+2d) must be nonmonotone. Since first term 1 is min, to avoid increasing, middle must be > last: P(n+d) > P(n+2d). For d=2 gives P(n+2)>P(n+4), d=4 gives P(n+4)>P(n+6), etc. Infinite decreasing positive integers impossible. Thus no 3-free permutation (sequence) of a well-ordered set N indexed by N or Z? If domain Z and range N (bijection), yes impossible. Therefore cannot split Z values into positive and nonpositive parity blocks and recursively use 3-free positive sequence. But for Z range, no well-ordering, so possible. Can we explicitly construct a 3-free permutation of Z? The finite recursive split into lower/upper halves cannot be infinite because each half would need to be 3-free permutation of a set of integers unbounded both below and above. We can recursively split Z into two subsets that are each order-isomorphic to Z and interleaved? For a bi-infinite discrete order, we can split into even/odd positions (in the permutation order) rather than value blocks. Maybe define a permutation q recursively on indices: q(2n) = 2 q(n) ? q(2n+1) = 2 q(n)+1? That's identity? Need ensure 3-free. Let's attempt to construct q:Z->Z satisfying σ(n+d,d)=-σ(n,d). This is a system of sign equations. Perhaps a simple formula exists: q(n) = n + (-1)^n f(n)? Or q(n) = (-1)^n n? Let's test q(n)=(-1)^n n (with q(0)=0). Is it a permutation? Values: n=0 0, 1 -1, 2 2, 3 -3, 4 4,... range {0, ±even? Actually (-1)^n n gives 0,-1,2,-3,4,-5,... not all integers (missing 1, etc.) Not permutation. q(n)=(-1)^n (2n+1)? Range odd signs, missing evens. Maybe q(n) = (-1)^n (n + c) no. Need a bijection. Perhaps q(n) = n + (-1)^n floor(n/2)? Let's not guess. Could use a "Hilbert hotel" to make a discrete order from the 2-adic recursive order. The finite bit-reversal order on [0,2^m) can be extended to a permutation of Z by choosing a bi-infinite sequence of finite blocks that are 3-free internally and with cross-block APs handled by magnitude. For example, arrange values in blocks of size 2^m with large gaps, so APs crossing blocks are monotone? Wait we want avoid monotone, not create. We can maybe make cross-block APs nonmonotone by using alternating signs/blocks. Alternative: Use a known construction: q(n) = (-1)^n * 2^{v_2(n)} * something? Let's explore sign equations. We need q:Z->Z bijection. For every d, σ(n,d) alternates in n. In particular, for fixed d, q(n+2d)-q(n) has sign constant on each parity class modulo 2d? More exactly σ(n+2d,d)=σ(n,d). Thus for each d, the relation "q(n+d) > q(n)" depends only on n mod 2d. For d=1, sign depends on n mod2. For d=2, sign depends on n mod4, etc. Can we define a total order on Z such that for each d, the order of points in each 2d-class is alternating? This resembles an ultrametric order. Maybe define order by the 2-adic valuation of the difference from a base? Consider the "2-adic order" on Z: say x <_2 y if |x-y|_2 is small? The 2-adic integers have a natural ultrametric tree; the order by lex of binary expansion is dense? Let's test on finite Z/2^m. Bit-reversal order on Z/2^m may be related to ordering by 2-adic intervals. For Z, perhaps use the 2-adic integers as a linear order? But need discrete type Z. Wait, a permutation of Z is not required to have order type Z if we define "permutation of Z" as a bijection π:Z->Z but the sequence is indexed by N? Usually a permutation of a set is just a bijection, no order type. But when we talk about a monotone AP in a permutation, we need an order on the permutation: either the domain order (usual Z or N) or the value order. If it's a bijection π:Z->Z, the "permutation" could be considered as the sequence (π(n))_{n∈Z} indexed by Z, so order type Z. If it's a bijection of the set Z (no specified sequence), "monotone AP in the permutation" might mean there exist x_i (values) such that π(x_i) is monotone and x_i is AP in the usual order. Then the domain order is usual Z (type Z), and the range values are compared by usual order. That's exactly q=π with indices values. So order type of domain is Z. Yes. But perhaps a "permutation of Z" in additive combinatorics is a bijection a:Z->Z, and a "monotone AP" means a(x_i) is monotone where x_i is an AP in the domain's usual order. So domain order fixed usual Z. That's what we've been doing: q=π. The sequence indexed by Z in usual order. So order type Z fixed; no freedom to choose a different discrete order. We are constructing π(n) values. The sign equations are correct. Can we solve sign equations with a bijection? Maybe use π(n)=n? No. Need signs σ(n,d) = sign(π(n+d)-π(n)). The equations only constrain signs, not magnitudes. If we can assign a total order (ranking) to integers such that sign(π(n+d)-π(n)) equals prescribed alternating pattern and π is bijective to Z. The sign pattern could be σ(n,d)=(-1)^{?} with ? maybe floor(n/d) parity. For consistency, maybe choose π(n) = (-1)^n n? But range issue. Could choose a bijection whose order (ranking) is given by a total order <_π on Z; we just need a total order of type Z? Wait π:Z->Z bijection: the order of values by usual integer order is the natural order on Z. The sequence π(n) in domain order is a permutation. The sign σ(n,d) is sign of usual difference between values π(n+d) and π(n). We cannot arbitrarily assign a total order; values are fixed integers. But any bijection induces some ordering of the domain points by value; this ordering is a total order of type Z? The set of domain points Z ordered by π-value: n <_π m iff π(n)<π(m). Since π is a bijection to Z, this order is order-isomorphic to Z (discrete bi-infinite). Thus constructing π is equivalent to choosing a linear order <_π on Z of type Z (a permutation order) such that for all APs, middle is extreme. Then assign π(n)=rank_{<_π}(n) (with ranks Z). So we need a linear order of type Z on the set Z with the 3-free property. This is more flexible than a sequence of positive values; the order must be type Z, but the underlying set is Z. The finite bit-reversal orders are of type finite; infinite recursive order maybe of type Z? Let's see. Can we define a linear order < on Z of type Z such that for every 3-AP a0 such that n0+d, n0+2d exist. The proof used sequence P_m indexed by m≥1 to the right in domain, not in order. It required P(n+d) > P(n+2d) for all d, where P is rank (range N). This is possible only if infinite decreasing positive ranks to the right in domain. Since domain has infinite right side, contradiction. Thus no 3-free ordering of type ω (or ω*) on Z? For type ω*, there is a last element (maximum rank), use left endpoint to get infinite increasing negative? If ranks nonnegative, maximum finite? Type ω* is order-isomorphic to N reversed; it has a maximum (last) but no minimum; assign ranks maybe 0,1,... in reverse? There is a global maximum rank? If order type ω*, the order has a last element with largest rank? If we map to Z ranks, any one-sided order has an endpoint with extreme rank? Actually type ω* embedded in Z as negative integers ..., -2,-1: has maximum -1 but no minimum; ranks unbounded below. The endpoint (maximum) as endpoint of domain APs may force infinite increasing ranks to left? Since ranks unbounded below? Let's analyze later. But likely one-sided blocks impossible for 3-free on Z because domain extends both ways and range one-sided has well-ordering obstruction. Therefore a 3-free order of type Z on Z may not exist? Because to be type Z, it must have a cut into lower ω* and upper ω (between adjacent elements). Consider the adjacent pair in the order (two consecutive ranks). The lower initial segment is type ω*, upper type ω. Are these initial/final segments convex in the order. Do 3-free conditions force something about parity across this cut? Not necessarily. But maybe use the global adjacent pair? Let's see. In any order of type Z, there is an adjacent pair? Yes, every element has immediate successor/predecessor. Let a <_π b be adjacent (ranks r, r+1). Does 3-free property impose that no other element lies between them (true). Could derive contradiction using APs with a or b as middle/endpoints? Since a,b adjacent in value order, any triple with one endpoint a and middle b? For AP in domain x,y,z with y middle. If values of x and z are on opposite sides of the cut? Hmm. Maybe any 3-free order of Z (under usual AP structure) cannot be discrete? If true, largest k=3. Let's test: Finite 3-free orders exist; infinite 3-free orders might be dense or non-discrete (like bit-reversal limit). Since a permutation requires discrete, maybe no 3-free permutation of Z, so every permutation has monotone 3-AP. Then need check if can avoid monotone 4-AP. Maybe answer 3. This seems plausible and nontrivial. The problem asks largest k; answer likely 3 (every permutation has monotone 3, but not 4). Need verify. Let's try prove every permutation of Z has a monotone 3-term AP. We found finite 3-free permutations, so not trivial by compactness; discreteness matters. Need proof. Then construct permutation of Z with no monotone 4-term AP. That would be a nice result. Is there known? "Every permutation of the integers contains a monotone 3-term arithmetic progression, but not 4." This rings a bell! There is a theorem: For any permutation of Z, there is a monotone 3-term AP. And there exists a permutation with no monotone 4-term AP. The largest k is 3. This might be the intended answer. Let's investigate. I recall a paper "Monotone arithmetic progressions in permutations" maybe by? It states: Every permutation of the positive integers contains a monotone 3-term arithmetic progression? But we proved no 3-free permutation of N? Actually for N one-sided, maybe every permutation of N contains monotone 3-AP (we sketched proof using global minimum? Wait that proof assumed no monotone 3-AP and derived contradiction using global minimum value 1 as endpoint in domain to right. That seems to prove every permutation of N has monotone 3-AP! Is that true? Let's test finite 3-free permutations of [n] exist for n=8, but infinite permutation of N cannot be 3-free. Yes, that matches. For Z, domain two-sided and range two-sided; global minimum absent, so 3-AP may be avoidable? But maybe discreteness still forces 3-AP. Need prove or disprove. Let's test potential 3-free permutation of Z with a simple construction. Could we use π(n) = n + (-1)^n? Not bijection? π(n)=n for even, π(n)=n-1? no. Try to find explicit 3-free π:Z->Z. If exists, answer 2. Let's attempt to construct using sign equations. Maybe there is a simple formula: π(n) = (-1)^n (n+1)/2 for even/odd? Let's test. Define π(2m)=m, π(2m+1)=-m? Range? even m∈Z gives all Z, odd gives nonnegative? not bijective. Need cover Z once. Maybe π(n) = floor(n/2) for even, -floor((n+1)/2) for odd? Let's list: n: ... -3,-2,-1,0,1,2,3,4 Even n=2m -> m: ... -2,-1,0,1,2 Odd n=2m+1 -> -m? n=-3 (m=-2)->2; -1(m=-1)->1; 1(m=0)->0; 3(m=1)->-1; 5->-2. Range duplicates 0,1,2? Not. Need a bijection; maybe use two interleaved copies: even positions get positive, odd get negative as above but impossible for 3-free due well-order? But we only need 3-free? We proved no 3-free sequence of positive integers indexed by Z, so that split cannot work. But maybe even positions get a 3-free sequence of integers unbounded both ways, odd positions get another, with no global sign separation. Need a 3-free permutation of Z restricted to each parity, and odd d condition handled by some alternating sign pattern not absolute separation. Let's try to solve sign equations locally. For d=1, choose pattern σ(n,1)=(-1)^n (q(n+1)-q(n) sign). Thus q has local extrema. Let’s denote even indices local minima, odd local maxima (WLOG). For d=3, σ(n,3) alternates with period 6. It might be related to d=1 signs? For a triple with d=3, endpoints n and n+6 same parity, middle n+3 opposite. Since odd local maxima/even minima, if endpoints even (min) and middle odd (max), middle is likely > endpoints? Not necessarily; a local max can be lower than some other even minima. But if we arrange magnitudes by parity scale (all odd maxima larger than all even minima), then d odd works. That was sign separation. If not, need more intricate. Could use a "layered" construction with values in intervals corresponding to dyadic scales, ensuring for each d the middle has larger scale. For d odd, endpoints same parity, middle opposite. If we assign scale based on parity (e.g., values with odd index have magnitudes larger than even index for d=1), but for d=3, endpoints index n,n+6 same parity, middle n+3 opposite; same parity scale works for all odd d. Thus absolute sign separation by parity handles all odd d. Then even d reduces to parity subsequences. This recursion suggests a 2-adic tree of scales. To make a permutation of Z, at each scale we need to split the index set into two cosets mod 2^r and assign them to value intervals? For finite, yes. For infinite, maybe we can assign to value intervals that are unbounded both ways at each scale? For parity separation, we need all values for one coset to lie on one side of all values for the other coset. In a permutation of Z, can we partition Z (domain) into two cosets (evens, odds) and assign their values to two subsets A,B of Z such that A and B are separated (all AB) and each of A,B is 3-free order-isomorphic to Z? As argued, if A and B separated in Z, one is an initial segment of type ω* and the other final of type ω (or finite). They cannot both be order type Z. But maybe the value subsets need not be separated globally; they only need that for any AP with odd d, the middle value is extreme relative to endpoints. This does not require all values of one parity to be on one side globally; it suffices that for each pair of endpoints of a given parity, the opposite parity middle is outside their value interval. Could be achieved by a "coloring" of values such that no two same-color values straddle an opposite-color value in a certain pattern? For d=1, adjacent endpoints in domain (n,n+2) same parity; middle n+1 opposite. Need middle extreme relative to those two endpoints for every n. This is a local condition, not global separation. For all odd d, stronger. Maybe a 2-coloring of integers (values) with domain parity can satisfy: For any n,d odd, value at n+d is outside interval between values at n and n+2d. This resembles a "2-free" property. Could be solved by assigning values according to 2-adic order of n: the parity of n determines the first bit of the rank; the rank order is a binary tree interleaved discretely. Let's attempt to construct a 3-free order of type Z on Z via a recursive "alternating" order, not block separation. For a set S indexed by Z, define an order <_S of type Z such that for any AP x,y,z in S (with S affine copy of Z), y is extreme. We want S=Z. This is like a self-similar "3-free" order. Does such an order exist? It would be a self-similar order of type Z. The finite bit-reversal orders are self-similar but finite. Infinite self-similar orders of type Z might exist using a "substitution" that shifts the center. Consider the usual order on Z: ... -2 < -1 < 0 < 1 < 2 ... . Can we transform indices by a bijection T:Z->Z such that the order <_π is T^{-1} of usual order? We need T to be a "3-free permutation" itself. We're looking for T. Maybe T(n) = v_2(n) * (-1)^n? Let's test small. Let's brute search for a small pattern to see if infinite possible. Suppose we want a permutation q of Z (values) with no monotone 3-AP. We can try to build around 0 recursively. Let positions (domain) integers. Choose q(0)=0 (WLOG by value translation? We can translate values). Need for all d: q(d) and q(2d) such that q(d) extreme between q(0),q(2d). Since q(0)=0, this means q(d) and q(2d) have same sign? If q(d)>0, then to be extreme (max) need q(2d)q(2d) (middle max) or q(d)<0 (middle min). If q(d)<0, nonmonotone if q(2d)>q(d) (middle min) or q(2d)>0. So condition: q(d) and q(2d) not in order 00 or both <0. If endpoints opposite signs, 0 is between -> monotone? Let's check triple (-d,0,d) values a,0,b. Monotone if 0 between a,b (a<00, q(-d) and q(d) have same sign. Also from q(0)=0 endpoint condition with d: q(d) and q(2d) same? Let's see for triple (0,d,2d): 0 middle? No 0 endpoint, d middle. Nonmonotone if q(d) extreme relative 0 and q(2d). This implies q(d) and q(2d) have same sign? If q(d)>0, need q(2d)0, q(n) positive, q(-n) positive? Then positive values used in pairs, negative none? Need both signs for unbounded both. Could have q(n) and q(-n) one positive, one negative? But triple centered at 0 requires same sign. Thus for each d, q(d) and q(-d) same sign. If sign positive for some d, both positive; if negative, both negative. So positive and negative values come in symmetric pairs. This is possible: assign pairs of same sign to ±d. Now consider d=1: q(1),q(-1) same sign. Suppose q(1),q(-1)>0. Then q(0)=0 is min in triple (-1,0,1). For triple (0,1,2), q(1)>0; need q(1) extreme: q(2)0 endpoint? Values q(-2),q(-1),0. To avoid monotone decreasing? q(-1)>0. If q(-2)>q(-1), sequence decreasing? q(-2)>q(-1)>0: monotone decreasing (bad). If q(-2)<0? q(-2)<0q(-1) (then q(-2)>q(-1)>0 decreasing). So need q(-2)0; endpoints q(-1)>0, q(3). To avoid monotone? If q(3)>q(1)>q(-1) maybe increasing? Need general. This seems complex. Maybe there is a known explicit 3-free permutation of Z: π(n) = n + (-1)^n ⌊n/2⌋? Let's test small with computational reasoning manually? Could try to find a periodic pattern of signs and assign magnitudes via a bijection. Let's search memory of "3-free permutation of integers". I recall a construction: Let p(n) = (-1)^n 2^{v_2(n+1/2)}? Not. Another angle: The problem may be from a contest where answer is 3 with a simple proof. Let's attempt to prove every permutation of Z has a monotone 3-AP. If true, construction avoiding 4 maybe also simple. Let's see if we can prove no 3-free permutation of Z. Assume π:Z->Z has no monotone 3-AP. For all n,d, π(n+d) is not between π(n),π(n+2d). In particular, for d=1, π is zigzag. So there are two cases: even positions local minima, odd local maxima, or vice versa. WLOG even minima, odd maxima: π(2m+1) > π(2m),π(2m+2); π(2m) < π(2m±1). Now consider the even subsequence E_m=π(2m). For d=2, E is zigzag: for all m, E_{m+1} is extreme relative E_m,E_{m+2}. Thus E has local maxima/minima alternating. Similarly odd O_m=π(2m+1) zigzag. Can we use the fact that the whole sequence is a permutation of Z (unbounded both directions) to derive contradiction? Maybe use extremal values in intervals? Since no global min, well-ordering not available. But perhaps use the alternating local extrema to find an infinite monotone AP of length 3 by a Ramsey argument on magnitudes. Let's denote local minima at even positions. There are infinitely many local minima and maxima. Since values are integers unbounded both ways, the set of local minima has no minimum/maximum. But we can pick a local minimum that is very low, and adjacent maxima very high? Maybe use d=2 to force a decreasing/increasing sequence of same parity that is monotone over distance 2? Already zigzag, not monotone. Need 3-AP, not 2. For a 3-AP to be monotone, the middle value must be between endpoints. We require middle extreme for all d. This is very strong. Maybe impossible due to transitivity of order on three parity classes? Let's use order relations. Let <_π be order by value. Condition: For all aZ. d=1 condition: q is zigzag. Thus in the order sequence, even/odd indices are alternating high/low. Can a zigzag bi-infinite sequence of distinct integers have the property for d=2? That means even subsequence and odd subsequence zigzag. So the order sequence is a 2D zigzag. This resembles a "2-adic zigzag" sequence. Is there a known sequence of all integers with this property? It might be the "van der Corput sequence" or "Stern-Brocot"? Let's try to construct values q(n) (ranks) satisfying sign pattern only; then we can set q(n) as some bijection to Z. The sign pattern for all d may be determined: maybe σ(n,d)=(-1)^{floor(n/d)}? Let's test consistency. For d=1, σ(n,1)=(-1)^n. For d=2, need σ(n,2)=? Condition for d=2: σ(n+2,2)=-σ(n,2). Could choose σ(n,2)=(-1)^{n/2?} for n even? But σ defined for all n. Maybe σ(n,d)=(-1)^{⌊n/d⌋}? For d=1 gives (-1)^n. For d=2: σ(0,2)=1, σ(1,2)=? floor(1/2)=0 ->1, but condition only relates n and n+2: σ(1,2) can be independent from σ(0,2)? Actually for d=2, sign alternates as n increments by 2, but no relation between residues mod 2. So σ(n,2) can be arbitrary for n even and n odd separately, with period 4? Wait σ(n+2,2)=-σ(n,2), so for even n, σ(0,2),σ(2,2)=-σ(0),σ(4)=σ(0), period 4 in n; for odd n separate. Thus many choices. But signs must come from a total order. There are global consistency constraints: For any n,d,e, the signs between n, n+d, n+2d, etc. must be transitive. For example, if q(n+d)>q(n) and q(n+2d) xZ directly using a greedy algorithm: enumerate domain indices in some order, assign distinct integer ranks, ensuring local 3-free conditions. Since constraints are local and infinite in both directions, maybe possible. Let's search for explicit known construction of a 3-free permutation of Z. Maybe "Thue-Morse sequence" gives signs, and then values can be assigned by a simple formula: π(n) = (-1)^n (n + t(n)) where t is Thue-Morse? Let's test. Suppose we define q(n) = (-1)^n (n + c) for n? Not bijection. But we can use a pairing function: For n even, assign a positive/negative value based on m; for n odd, assign the other. Need avoid 3-free obstruction if one parity values all positive. But if both parity subsequences are permutations of Z (unbounded both), and for odd d we rely not on global sign but on a sign pattern: q(2m) and q(2m+1) maybe have alternating signs depending on m such that for any odd d, middle extreme. What if q(2m)=A_m, q(2m+1)=-A_m, with A_m a permutation of positive? We proved impossible because A_m (m∈Z) positive no 3-free. But if q(2m)=A_m, q(2m+1)=B_m where A_m,B_m are permutations of Z and for odd d, the middle is extreme due to some relation between A and B. For d=1: q(2m+1)=B_m must be extreme between A_m,A_{m+1}. For all m. For d=3: q(2m+3)=B_{m+1} extreme between A_m,A_{m+3}, etc. Hard. Maybe choose A_m and B_m interleaved such that B_m is always outside interval [A_m,A_{m+1}] for d=1, and B_{m+e} outside [A_m,A_{m+2e}] for odd d. This resembles two sequences A,B that are "2-free" mutually. Could use A_m = m, B_m = M - m? For finite. For infinite, maybe A_m = m, B_m = -m? Let's test q(2m)=m, q(2m+1)=-m. Range? m∈Z: even gives all Z, odd gives all Z (duplicates). Not bijection. Shift: q(2m)=2m? q(2m+1)=? Need disjoint ranges. Use q(2m)=2m, q(2m+1)=2m+1? identity bad. Maybe q(2m)=2m, q(2m+1)= -2m? duplicates. Use a pairing function that maps pairs (m, parity) to Z bijectively and has some order property. For example, q(2m)=2m, q(2m+1)=2m+1? no. q(2m)=m (all integers), q(2m+1)=? Need remaining none. We need partition Z into two sequences A_m,B_m. A natural partition: A_m=2m (evens), B_m=2m+1 (odds). q(2m)=2m, q(2m+1)=2m+1 identity. Could rearrange within partitions: A_m=2a(m), B_m=2b(m)+1 with a,b permutations of Z. For d=1, triple (2m,2m+1,2m+2): values 2a(m), 2b(m)+1, 2a(m+1). Need middle extreme. This is a condition on two permutations a,b of Z. For d odd, similar. For d even, reduces to a and b individually 3-free? If d=2e: (2m,2m+2e,2m+4e): values 2a(m),2a(m+e),2a(m+2e); need a no monotone 3-AP. Similarly b. Thus we need two 3-free permutations a,b of Z such that cross conditions hold. This is recursive. Maybe take a=b=identity? Then d=1: 2m,2m+1,2m+2 monotone (bad). Need cross. Could take a(m)=m, b(m)=-m? Check d=1: values 2m, -2m+1, 2m+2. For m=0:0,1,2 monotone (bad). b(m)=M-m finite. What about a(m)=m, b(m)=m+1/2? Not. Maybe use a(m)=m, b(m)=m but shift by large alternating? Hmm. Let's search memory of explicit 3-free permutation of Z: There is a sequence called "3-free permutation of the integers" given by a(n) = (-1)^n floor(n/2)? Let's test some candidates. Candidate: q(n) = (-1)^n * (n + |n|?)/2? For n≥0: even 2m -> m, odd 2m+1 -> -(2m+1)? Range duplicates? n=0 0,1 -1,2 1,3 -3,4 2,5 -5,... range includes all odd negative and nonnegative? Missing negative even. Add negative n? Could define for negative n to cover remaining. For example: For n≥0: q(2m)=m, q(2m+1)=-(2m+1). Range nonnegative (m≥0) and negative odd. For n<0: need cover negative even. Let n=-2m-1? q? Maybe q(-2m-1)=-(2m+2)? and q(-2m-2)=? This could be bijection. Let's construct: List n: ..., -4,-3,-2,-1,0,1,2,3,4 Want q: maybe ...? Use q(n) = (-1)^n ceil(n/2)? Let's compute q(n)=(-1)^n floor((n+1)/2)? n=0:0,1:-1? floor1? no. Try define q(n) by taking the sequence of integers in order: 0, -1, 1, -3, 2, -5, 3, -7, 4, -9,...? This is q(0)=0, q(1)=-1, q(2)=1, q(3)=-3, q(4)=2, q(5)=-5,... plus negative indices maybe continue symmetrically: q(-1)=? Need bijection. This sequence for n≥0 uses nonnegative and negative odd. It zigzags? 0,-1,1,-3,2,-5: signs -, +, -, + yes. d=1 no monotone triples? 0,-1,1: middle -1 min; -1,1,-3: middle 1 max; 1,-3,2: middle -3 min; good. d=2: subsequence even: 0,1,2,3,... monotone! Bad (0,1,2 monotone). Need even subsequence zigzag too. Modify by recursive: q(2m)=? should be a 3-free permutation of some subset. If even positions use all nonnegative integers in natural order, bad. Need even positions 3-free. Maybe use q(2m) = 2m? no. Let's attempt recursive construction on Z with a "center" that moves. For finite bit-reversal on [0,2^m-1], the order (sequence) has a certain center. To extend to Z, we can use a bi-infinite sequence of blocks: ... B_2, B_1, B_0, B_1, B_2 ...? Where each block is a scaled 3-free permutation, and values in different blocks separated by magnitude to handle cross-block APs. For example, arrange value intervals I_m = [L_m,U_m) with large gaps, and within each interval use a 3-free pattern. Cross-block APs (indices in different blocks) may have middle in a block of larger magnitude than endpoints, making it extreme. If we can make the block containing the middle index have values much larger/smaller than endpoints for any AP, then no monotone. Domain Z can be partitioned into intervals (blocks) J_m of sizes growing, e.g., J_m = [ -2^m+1, 2^m-1 ]? Nested. For any AP, the middle index lies in a smaller/nested block? If blocks are centered at 0, an AP crossing blocks: middle may be in a block with smaller magnitude than one endpoint? Not guaranteed. Alternatively partition domain into dyadic intervals at each scale, like a binary tree over Z. For finite [0,N), bit-reversal uses a complete binary tree; the middle of any AP lies in a node at a scale where its value block is outside endpoint blocks. For infinite, use a binary tree on Z that is locally finite and covers Z with no boundary. Maybe possible via the 2-adic integers: the domain Z is a subset of the 2-adic integers; the finite bit-reversal order on Z/2^m corresponds to a total order on the 2-adic cylinder sets. The limit is an order on the 2-adic integers (not Z) that is 3-free. We need restrict to the ordinary integers Z and then map to a discrete permutation? The 2-adic order restricted to Z might be a 3-free order, but not type Z? Maybe we can re-index it to Z by a monotone bijection? Wait the order itself on Z (values ordered by 2-adic lex) may be dense/non-discrete, but the set Z is countable; any countable order can be enumerated in a sequence of type Z, but the enumeration order would be different from the 2-adic order. We need the sequence order to be the 3-free order. If the 3-free order on Z is not type Z, cannot be a permutation of Z. But perhaps the 2-adic order restricted to Z is actually type Z? Let's check. Define a linear order on Z by the 2-adic metric: x <_2 y if x-y is divisible by a higher power of 2? For finite mod 2^m, bit-reversal order maybe corresponds to ordering by the reversed binary digits, which is a cyclic order? Let's examine order on Z_2: For integers mod 2^m, bit-reversal order 0,4,2,6,1,5,3,7. This order has adjacency between 6 and1? In mod 8 cycle? The sequence 0-4-2-6-1-5-3-7. If wrap around (7 to 0) maybe it's a cyclic order? Differences: 4,2,4,3,4,2,4,1? Not constant. It might be the order of the 2-adic integers by the Gray code? The infinite limit might be order type of the 2-adic integers (not order type Z). Let's test 2-adic order on Z: Define x 1< -2? sequence 1,2,-2 non? middle 2 max. AP 0,1,2: 0,1,2 monotone! Bad. Maybe order by 2-adic valuation descending? 0 last? Let's look for proof of no 3-free permutation of Z. Suppose order <_π type Z. Choose a point a. Since order type Z, there are elements immediately below and above a. Maybe use AP homothety centered at a: For any d, b=a-d, c=a+d. Condition says a is extreme relative b,c; so b,c are on the same side of a in the order. Thus for each d, the two points symmetric around a lie on the same side of a. Let S^+ = {x: x >_π a}, S^- = {x: x <_π a}. Then for every d, a-d and a+d are both in S^+ or both in S^-. This is a strong symmetry: the sets S^+ and S^- are "2-symmetric" around a: x∈S^+ iff 2a-x∈S^+. Since S^+ is a final segment of the order (type ω) and S^- initial (type ω*). Can a final segment of type ω be symmetric under reflection about a? Let's analyze. Translate a to 0. We have a partition of Z\{0} into P (positive in order) and N (negative in order) such that for all d>0, d and -d are both in P or both in N. Thus the set A = {d>0: d∈P} and B = {d>0: d∈N}; partition of positive integers. P = A ∪ (-A), N = B ∪ (-B). In the order, all elements of P are greater than 0, all N less than 0. Since the order type is Z, P is a final segment of type ω (has a least element >0 in order) and N is an initial segment of type ω* (has a greatest element <0). But the sets P,N as subsets of Z are symmetric. Is that impossible? Not obviously; e.g., P = even nonzero, N = odd nonzero is symmetric, but not an initial/final segment of a Z order? We can order N as ...? If P is final segment type ω, it has a least element in the order, call p. Then all elements greater in order are in P; N are below 0. The actual integer values of P,N can be arbitrary. Symmetry just says if p∈P then -p∈P, etc. Now consider the least element p of P (immediate successor of 0 in order). Its negative -p is also in P. Since p is least in P, -p is ≥_π p (i.e., p <_π -p unless equal, impossible). So in the order, 0 < p < -p. Now p and -p are symmetric around 0. Consider the AP with middle? p and -p endpoints? Their midpoint is 0, which we already used: 0 is extreme (0 < p < -p? Wait values in order: 0 < p < -p. For AP (-p,0,p), middle 0. In order, 0 is less than both p and -p, so extreme (min). OK.) Now use the immediate predecessor of 0 in order, call q∈N (greatest negative). Symmetry: -q∈N. In order, -q < q < 0? Since q greatest in N, -q <_π q. Can we find a 3-AP violating? Consider p and q? Maybe their difference? p and -q? Hmm. The symmetry at a single point is necessary but not sufficient. It may hold for any 3-free order: for every a, the partition around a is reflection-symmetric. This is a strong condition: For all a, the order <_π is such that for every d, a-d and a+d are on same side of a. Equivalently, for any a 0<1. (1,2): diff1 even =>1<2. Then 0<1<2, bad for AP 0,1,2 (middle 1 between). So not 3-free. Maybe x1>2? For 0,1,2: 1<0? diff1 odd =>1<0; 2<1; so order 2<1<0, middle 1 between (decreasing), bad. Need middle extreme, so for AP x,y,z, the valuations of differences should make y not between. If xy and y>z). We want not. For pair differences: y-x=d, z-y=d. The order relation between x and y and y and z should be opposite (so y extreme? Let's check: If xz, y max; if x>y and yz OR x>y and y0 usual, which is not 3-free. So order cannot be translation-invariant. Maybe use a "lexicographic order" on binary expansions of n (two-sided). For finite bit-reversal, order of values q(i) is bit-reversal; equivalently, i <_π j iff bit-reversal(i) < bit-reversal(j). For infinite, if we take the 2-adic integer i (including negative via 2-adic expansion) and reverse its infinite bits? There is a natural bijection from 2-adic integers to 2-adic integers by bit reversal? It may map Z to a subset, and the usual order on 2-adic? Not. Let's consider using the 2-adic integers. The finite bit-reversal permutation on Z/2^m can be seen as a group automorphism? Actually map φ_m: Z/2^m -> Z/2^m by reversing m bits. For x∈Z, its residue mod 2^m maps to φ_m(x mod 2^m). As m→∞, this defines a map φ: Z_2 -> Z_2 (2-adic integers) by reversing the infinite 2-adic expansion (from left to right?). The usual order on finite images 0..2^m-1 (natural order) corresponds to some order on Z_2? The limit order on φ_m(i) as m→∞ might be the 2-adic order? If φ maps ordinary integers Z (embedded in Z_2) into Z_2, the order induced by the natural order on 2-adic? But we need map to Z, not Z_2. Maybe there is no 3-free permutation of Z, and the 2-adic limit order is non-discrete. Let's try to prove no 3-free order of type Z using the symmetry around each point and discreteness. Let < be a 3-free order of type Z on the set Z. For each a, define S_a^+ = {x: a<_x} (final segment). Reflection symmetry: x∈S_a^+ iff 2a-x∈S_a^+. Since S_a^+ is type ω (has a least element s(a) immediate successor of a). Let s(a) be the immediate successor of a in the order. Then -? 2a-s(a) is also in S_a^+, so s(a) <_π 2a-s(a) (or equal if s(a)=a, no). Thus the reflection of the immediate successor is strictly above it. Similarly, the immediate predecessor p(a) has reflection 2a-p(a) strictly below p(a). Can we derive a chain of immediate successors? Let a_0=0. Let a_1 = s(0), a_{-1}=p(0). Symmetry: -a_1 ∈ S_0^+, so 0 < a_1 < -a_1. Also -a_{-1} < a_{-1} <0. Now consider point a_1. Its immediate successor s(a_1) is in S_{a_1}^+. Symmetry around a_1: reflection of s(a_1) is 2a_1-s(a_1) also in S_{a_1}^+, and s(a_1) < 2a_1-s(a_1). The actual integer values? a_1 and -a_1 are symmetric around 0. Maybe a_1 is an immediate successor of 0, so no element between 0 and a_1 in order. But what about the AP (0, a_1, 2a_1)? The middle a_1 must be extreme relative 0 and 2a_1. Since 0 < a_1, to be extreme (max), we need 2a_1 < a_1 in order. Thus 2a_1 ∈ S_0^- (below 0). But symmetry around 0 says if 2a_1 is below 0, then -2a_1 is above 0. OK. So for each immediate successor a of 0, 2a is below 0. More generally, if x is immediate successor of y, and y < x, then for AP (y,x,2x-y), middle x must be max, so 2x-y <_π y. Thus the "doubling step" from y to x jumps across y downward. This resembles a recursive tree: immediate successors map via homothety to predecessors. Could discreteness + this lead to an infinite descending chain of integers that forces a well-ordering contradiction? Let's follow from 0: let a=s(0)>0 in order. Then 2a < 0. Let b = immediate predecessor of 0 (call p<0 in order). Then -2p >0? Since AP (p,0,-p)? Wait p is immediate pred of 0. For AP (2p, p, 0) with middle p? Actually p,0? Use endpoint 0? For AP (2p,p,0), middle p, endpoint 0. Since 0 > p (p<0 in order), to avoid decreasing 2p>p>0? We need p extreme. If p is min, need 2p > p? Not. Let's systematically use endpoint condition: If y < x in order (x succ y), then for AP (y,x,2x-y), middle x is max, so 2x-y < y. Thus the point 2x-y is an immediate? not immediate, but below y. This gives a mapping T(y,x)=2x-y below y. Starting at 0, a=succ(0) gives 2a below 0. Since 2a is an integer (could be large). It lies below 0, so in the initial segment. The initial segment below 0 is type ω* (order like ..., -2,-1). It has a greatest element p (immediate predecessor of 0). Since 2a < 0, 2a ≤_π p (i.e., p is above 2a). The integer value of 2a may be anything. Now apply to p? p is below 0, and 0 is its immediate successor? Actually p = pred(0), so p <_π 0 with no between. For y=p, x=0, since y 2x-y changes the 2-adic valuation; iterating produces an infinite set of points all below 0 with no greatest? But initial segment below 0 has greatest p. Could iteration produce a point below 0 greater than p? impossible if p greatest. Let's check: 2a is below 0, so 2a ≤_π p. If 2a ≠ p, then p is above 2a. Could the rule applied to some pair force a point below 0 that must be above p, contradiction? Not yet. Maybe use the fact that the immediate successor a is the least point above 0. The point -a is also above 0 (by symmetry), so a <_π -a. Thus a is not the greatest? It is least. The point 2a is below 0. What about -2a? Above 0. For AP (a, -a, -3a)? Middle -a? Since a and -3a endpoints? If a and -a are above 0, their midpoint 0? Not. Let's choose actual integer values to see if a 3-free order could exist. Try small construction on a finite window around 0 of size, say, 9, and see if constraints force a pattern. We can perhaps do a backtracking mentally or derive that no infinite because of a local finite contradiction? But finite 3-free permutations exist on intervals, so no local contradiction of bounded size. The contradiction must use unboundedness/well-ordering or discreteness. Discreteness: In a 3-free order of type Z, each point has immediate successor/predecessor. Finite approximations can be extended to an infinite order that is not discrete (e.g., 2-adic limit). The property may force a non-discrete order (every cut has no adjacent pair) if infinite. Let's see if the symmetry condition implies no adjacent pairs. Suppose a <_π b are adjacent. Consider the AP with endpoints a,b? Need a,b same parity for midpoint integer. If they are same parity, midpoint m=(a+b)/2. Since a,b adjacent, m is either below a or above b. Condition for (a,m,b) (if aZ (permutation). We need f 3-free. Can we define f recursively using the usual order's cut at 0? For finite bit-reversal on [0,2^m), the order has a cut between the even and odd domain halves: all even domain elements (which in usual order are interleaved) before all odd. For Z, maybe we can split domain Z into two sets that are both bi-infinite but in the order one is initial segment of type ω* and the other final of type ω. For the order to be type Z, the split sets must be one-sided in the order, but their domain sets can be bi-infinite. For example, let lower block L = {even integers ≤0?} and upper block U = {odd integers ≥?}? Need all L before all U, with L having a maximum in order and U a minimum. But the domain sets L,U are not convex in usual order; APs crossing between L and U need condition. Maybe choose L = nonpositive integers, U = positive integers (usual cut). If all nonpositive before positive in the order, then the order is type? Nonpositive as a set in order can be arranged type ω* (has maximum), positive type ω (has minimum), so total type Z. Does this simple cut help 3-free? For an AP a lower. - If b nonpositive and endpoints positive? middle min (good). - If all three in same side, need recursive property within nonpositive or positive. - If AP crosses the cut but middle? Since the cut is at 0: possible a≤00 in upper block. For an AP a0), middle lower. Then middle is not necessarily extreme: order lower all lower, b lower < c. Monotone if a <_order b < c (increasing) or c < b < a impossible because c upper. So to avoid, need b not after a within lower; i.e., a >_order b (so b is min? order: b < a < c? Values lower: b and a, with a above b; sequence a,b,c: a > b, b < c => nonmonotone (middle min). If a <_order b, then a_order b. This is a cross condition. If a<0≤b_order c (middle max). Cross condition. If a<0 -x) we get a sequence of nonnegative integers indexed by one-sided domain with no monotone 3-AP. Does a one-sided permutation of N (domain N, range N) with no monotone 3-AP exist? We earlier used global minimum 1 and right endpoint to prove no if domain has infinite right side. For domain N, yes no. Thus no 3-free one-sided permutation of N. Therefore no 3-free order of type ω on N or ω* on nonpositive. So splitting at 0 into one-sided blocks impossible if each block must be 3-free for APs entirely within block. Thus a 3-free order of type Z cannot have a cut where both sides are one-sided and each side contains infinitely many APs? Wait the lower block nonpositive in usual order is one-sided (no positive), and APs entirely within it exist; they would require a 3-free one-sided order, impossible. Hence no such cut. More generally, in any order of type Z, there is a cut between adjacent elements p<_π q. The lower initial segment L={x: x< q?} maybe infinite one-sided in order, but as a subset of Z it may be two-sided in usual order (e.g., all integers except q and above in order). The usual domain subset L need not be one-sided, so APs within L may not reduce to one-sided integer order. The obstruction for one-sided range required the domain set to have an endpoint in the usual AP structure? We used domain has infinite right side from the minimum value point. If L as a subset of Z is not an interval but arbitrary, it might still have the global minimum point a∈L (in order) and APs within L extending infinitely in one direction? Need check. Suppose L is an initial segment of the 3-free order, of type ω* (if it's all below q). It has a greatest element p (immediate predecessor of q? if L below q). Let a=p be the maximum of L in order. For APs entirely within L with a as an endpoint in the domain usual order, do we get a contradiction like before? The range values of L are unbounded below (since type ω*), not well-ordered, so the positive decreasing sequence obstruction may not apply. If L type ω* (order ... < a_{-1} < a_0 < p), assign ranks -∞? There is a maximum rank 0 but no minimum; values unbounded below. A global maximum in order (p) as endpoint could force an infinite increasing sequence of ranks to one side? Ranks are negative integers (unbounded below), so increasing (towards 0) infinite? Let's analyze. For a one-sided order of type ω* on some subset S of Z, with a maximum element a (in order). If the domain S has infinite rays from a in the usual integer order, can we derive contradiction? The earlier proof for range N used global minimum (rank 0) and required for endpoint a: P(a+d) > P(a+2d) for all d in a direction, giving a decreasing sequence of positive ranks. For range ω* (ranks ..., -2,-1,0), global maximum rank 0. For endpoint a, to avoid monotone, if a is maximum rank, then for AP (a, a+d, a+2d) in domain, sequence ranks (0, r_d, r_{2d}) with r<0. Since 0 is max, monotone decreasing would be 0 > r_d > r_{2d}. To avoid, need r_d < r_{2d} (middle min? values 0, -5, -3: 0>-5<-3 non; if 0>-3>-5 decreasing bad). So for all d in a direction, r_d < r_{2d}, i.e., ranks (negative) increase to 0 along even subsequence: r_2 < r_4 < r_6 < ... (e.g., -10<-4<-1). This is an infinite increasing sequence of negative integers bounded above by 0, possible! e.g., -n. No contradiction. So one-sided order of type ω* on a two-sided domain might be possible. Similarly type ω on a two-sided domain with global minimum gave decreasing positive sequence impossible? Wait for type ω (ranks 0,1,2,... unbounded above), global minimum rank 0 gave r_d > r_{2d} (decreasing positive) impossible. For type ω* (ranks ...,-2,-1,0), global maximum rank 0 gives r_d < r_{2d} (increasing negative) possible because no lower bound? Actually negative integers have no minimum, so infinite increasing sequence bounded above possible. Thus asymmetry: type ω (well-order from below) impossible for range N with two-sided domain? But if the domain set S has infinite ray in the direction used. For type ω* possible. In a Z order, there is both an initial segment of type ω* (below a point) and final segment of type ω (above a point). The final segment of type ω might be impossible if its domain subset has an infinite usual ray from its least element. But the domain subset (all points above the cut) may not contain a full arithmetic ray from its least element? Let's see. Let q be the least element of the final segment U (immediate successor of p). U has a minimum in order, rank 0 (relative). If the domain set U contains, for some d>0, all points x+kd? The APs with x=q as endpoint need q+d, q+2d to be in U? The condition for 3-free only applies if all three points are in the domain Z (always), but the order relation of q+d and q+2d relative to q may not be both in U? Wait q is minimum of U, but q+d may be below q in order (i.e., in L) even though domain point is different. The endpoint proof used that q has global minimum rank among all points, not just U. If q is not global minimum of the whole Z order (there are L below), then for AP (q, q+d, q+2d), q may not be minimum among the three; q+d could be below q. The constraint is different. So the one-sided obstruction applies only to a point that is global minimum (or maximum) in the entire order, which doesn't exist in type Z. Thus possible. So no easy contradiction. Let's search for actual known result. I recall a theorem: "Every permutation of the integers contains a monotone 3-term arithmetic progression" might be false; there is a "3-free permutation of the integers" constructed by S. P. ? Maybe called "The 3-free permutation of the integers" exists and is unique up to symmetries? Let's think. There is a known concept "3-free ordering" of the integers. I remember reading about "A 3-free permutation of the integers" by? It might be possible and given by the "2-adic ordering". Let's search memory: A permutation a_0,a_1,a_2,... of the nonnegative integers is 3-free if no a_i < a_j < a_k for iZ given by f(n) = (-1)^n (n+1) for even? Let's test some candidate sequences from 2-adic order. The 2-adic order (also called "2-adic ordering" of the integers) can be defined recursively: List integers as: ... , -3, 1, -7, 3, -11, ...? Need a discrete order. Maybe start from 0 and define neighbors recursively: The immediate neighbors of 0 should be 1 and -1? But AP ( -1,0,1) requires 0 extreme, so -1 and 1 must be on same side of 0 in order. Thus they cannot be immediate predecessor and successor of 0; both must be on same side (both greater or both less). So 0 cannot have both 1 and -1 as neighbors; one of them must be far. For example, order ... < 1 < 0 < -1 < ...? Then 1 and -1 both? If 1<0<-1, endpoints 1,-1 both on opposite sides? For AP (-1,0,1) in usual order (-1,0,1): order values: 1 < 0 < -1. Middle 0 is between (bad). If order 1 < -1 < 0? endpoints 1,-1 both <0, 0 max (good). So -1 and 1 must be together below 0 or together above 0. Thus immediate neighbors of 0 are some pair ±a? Symmetry: if a above 0 then -a above 0. The least above 0, call a, and its negative -a is also above. So the two immediate? The order above 0 begins with a, then eventually -a (or -a then a? Symmetry only says both above; which is first? If a is least above 0, -a > a). So 0's immediate successor is a, and -a is somewhere above. The immediate predecessor p of 0 has -p below 0 and -p < p. Thus 0's neighbors are p (some integer) and a, with p and a not necessarily ± related? By symmetry around 0, if p below then -p below, not above. a and p independent except AP constraints. Let's try choose a=1. Then 1 is immediate successor of 0; -1 also above 0, so 0<1<...<-1? Actually 0<1< -1 (in order). AP (0,1,2): middle 1 max, so 2 below 0. Thus 2 < 0. Symmetry: -2 above 0. AP (-2,-1,0): middle -1? Values 2? Let's set order so far: 2 < 0 < 1 < -1, -2 above 0 somewhere. Need place -2. AP (-2,-1,0): usual -2,-1,0. Middle -1. We know -1 above 0; 0 above 2. Need -1 extreme relative -2 and 0. Since -1 >0, to be max need -2 < -1 (could be below 0 or above but less). AP (-1,0,1): endpoints -1,1 both above 0, middle 0 min OK. AP (1,2,3): middle 2? 2 below 0; endpoints 1 above, 3 unknown. Need 2 extreme. Since 2<1, to avoid decreasing (1>2>3) need 2<3 (middle min). So 3 >2. Many. Maybe we can build a tree. Choose a=1. Then 2 below 0. What about 4? AP (0,2,4): middle 2. 2 below 0. Endpoint 0 above 2. To avoid increasing? Values 0,2,4 with 2 maybe below 0. If 4 below 2, sequence 0 >2>4 decreasing bad; need 4 >2 (middle min) or 4 above? So 4 >2. AP (2,4,6): middle 4. 4 >2. Need 4 extreme relative 2,6: must be max, so 6<4. Thus even positive integers order: 2 < 4 < 6? Wait we got 2<4 and 4>6? Let's carefully: AP (0,2,4): values r0=0, r2, r4. r2<0. Nonmonotone: if r2 is min, need r2 < r4 (and r0> r2). So r4 > r2. (Could r4<0 or >0; just >r2.) AP (2,4,6): r4 is middle. Since r4>r2. To avoid increasing r2r6 (middle max). So r6r6>r8 need r8>r6 (middle min). So r8>r6. Thus even subsequence (2,4,6,8,...) zigzags: r2 < r4 > r6 < r8 > ... starting with min at 2? Also from AP (0,2,4) we only know r2-1>-3 decreasing! bad. So not. Let's search memory more concretely. There is a MathOverflow question "3-free permutations of the integers" with answer: Yes, they exist. One construction: ... , -6, 2, -14, 3, -26, 4, -2, 0, -12, 1, -24, 5, ... Maybe derived from the 2-adic order. The sequence might be indexed by the 2-adic integers in the usual order? Hmm. Let's try to formulate a recursive construction for a 3-free permutation of Z using a "central" point and two rays. We need an order of type Z. We can build it as a bi-infinite sequence (in order) of domain integers. Let the sequence in order be ... u_{-1}, u_0, u_1, ... (u_m is domain integer at order position m; q(u_m)=m). We need for any AP in domain (a,b,c with b=(a+c)/2), the order positions of b is not between positions of a,c. This is symmetric. We can try to construct the ordered sequence recursively by splitting the domain into two sets that will occupy the left and right rays. For finite bit-reversal on [0,2^m), the ordered sequence (by positions) is: 0,4,2,6,1,5,3,7 for m=3. It can be generated recursively: L_m = (L_{m-1} on evens) followed by (L_{m-1} on odds) in terms of domain? Actually ordered sequence S_m of domain elements: S_{m+1} = [2*s for s in S_m] followed by [2*s+1 for s in S_m]? Let's check S_3 from S_2? S_2 for [0,4): bit-reversal order domain: 0,2,1,3. S_3: evens (2*s): 0,4,2,6; odds (2*s+1):1,5,3,7. Concatenate: 0,4,2,6,1,5,3,7. Yes. So finite 3-free order of [0,2^m) is given by S_m = (2 S_{m-1}) ∪ (2 S_{m-1}+1) concatenated. For Z, maybe use a similar recursion but with a "shift" to keep order type Z: S_{m+1} = (2 S_m) ∪ (2 S_m + 1) but with S_m a bi-infinite sequence of type Z. If we can find a bi-infinite sequence S_0 of type Z and define S_{m+1} by interleaving scaled copies? But concatenation of two bi-infinite sequences is not type Z. Instead we can interleave the two copies in a way that preserves type Z, e.g., S_{m+1}(2n) = 2 S_m(n), S_{m+1}(2n+1) = 2 S_m(n)+1? That is identity-ish and not 3-free. Or S_{m+1}(n) = 2 S_m(n) + (n mod 2)? Need a permutation of Z. Wait, the finite recursion on domain indices (order positions) used concatenation because the order was finite. For infinite, we need a "self-similar" substitution on a bi-infinite word. This is like constructing a bi-infinite sequence over alphabet Z that is a permutation and satisfies a 3-free property. Maybe use a morphism that is a permutation of Z and preserves 3-free. For example, define a sequence q(n) recursively by q(2n) = 2 q(n), q(2n+1) = 2 q(n)+1? That's q(n)=n. q(2n)=2q(n)-1, q(2n+1)=2q(n)+1? Let's test. We want a bijection T:Z->Z such that the order sequence is T? Actually q is the permutation (value at domain index n). If q is defined by q(2n)=A(q(n)), q(2n+1)=B(q(n)) with A,B linear maps partitioning Z, perhaps 3-free property self-similar. Suppose we can find A,B:Z->Z (affine) such that A(Z) and B(Z) are disjoint and union Z, and for any AP, if d odd, the middle value (odd index) is extreme due to A,B separation; if d even, reduces to q on n. Finite used A(m)=2m, B(m)=2m+1 with A values lower block, B higher block (concatenation in order), not interleaving. For infinite, maybe choose A(m)=2m, B(m)=2m+1 but interleave their outputs in the order via q's index mapping? Wait q(2n)=A(q(n)) means the value at even domain index is A of value at n. The order sequence q(n) (domain index n) is being constructed; the parity of domain index determines which affine transform of q(n). The order positions are n (usual). If A and B have separated ranges (e.g., A even, B odd), then for d=1, triple (2n,2n+1,2n+2) values A(q(n)), B(q(n)), A(q(n+1)). Need B(q(n)) extreme between A(q(n)),A(q(n+1)). If A and B ranges are separated (all B > all A), then B is max, good. But ranges A=even, B=odd are not separated in Z (odds/even interlace). We need A(Z),B(Z) separated: e.g., A(m)=2m (even) and B(m)=2m+1 (odd) not separated. Could choose A(m)=2m, B(m)=2m+2? not partition. For partition of Z into two affine sets separated (one all less than other), impossible as both Z? Affine sets of form am+b with a≠0 are two-sided, cannot be separated. But they need not be affine; we can partition Z into two sets A(Z),B(Z) that are separated and both bi-infinite? As argued, two separated subsets of Z with all A all A, then even/odd values separated, good. For even d, even subsequence q(2n) values in A (one-sided) indexed by n∈Z must be 3-free. We proved impossible if A is well-ordered (N) because global minimum. If A is Z_{≤-1} (bounded above, type ω* in value order), maybe possible. So one range can be upper unbounded (N) and the other lower unbounded (negative). The upper range N for one parity would be impossible for 3-free if that parity subsequence is indexed by Z. Thus both parity subsequences cannot be one-sided; at least one would be N. But maybe choose A=Z_{≤-1}, B=Z_{≥1}, both one-sided (one below, one above). The subsequence on A (negative integers) indexed by Z might be possible? The subsequence on B (positive) impossible. So parity separation impossible. But we don't need full separation; perhaps use a partition into two sets that are interlaced but still ensure odd d via a sign pattern. Let's look at the sign equation approach. We only need signs, not magnitudes. Maybe we can define a total order < on Z of type Z directly via a "2-adic" discrete order. Let's search for a known discrete 2-adic order on Z. There is the "van der Waerden ordering" or "bit-reversal ordering" of the 2-adic integers. The 2-adic integers with the lex order of their binary expansion (from least significant to most?) might have order type Z? Let's examine. Represent each integer n in two's complement binary infinite to the left? For finite bit-reversal on [0,2^m), the order is by the reversed m-bit representation in natural order. Equivalently, the domain index n is ordered by the integer whose binary expansion is the reverse of n's m-bit expansion. As m grows, this order on Z_2 (2-adic integers) is: x < y if the reversed infinite binary expansion of x is less than that of y in the usual (binary) order? Reversing infinite binary expansions yields a map from 2-adic integers to ordinary real binary expansions? The natural order on reversed expansions is the usual lexicographic order from the least significant bit onward? This order on Z_2 might be dense? But restricted to ordinary integers (finite binary with leading zeros?) maybe discrete? Let's define for n∈Z, its 2-adic expansion (infinite to left) ... b_2 b_1 b_0. Reverse it to a real in [0,1): 0.b_0 b_1 b_2 ... (infinite to right). Order by this real. This is a total order on Z_2. Is it discrete? For ordinary integers, their 2-adic expansions are eventually periodic? Actually finite integers have finite binary expansion with leading 0s in 2-adic? In 2-adics, nonnegative integers have finitely many 1s, leading 0s; negative integers have eventually 1s. Reversing gives real sequences with tails 0 for nonnegative, tail 1 for negative. The order by reversed bits: nonnegative integers have reversed sequences with finite number of 1s then all 0; negative have finite prefix then all 1. This order might have type Z? Let's test on small integers. Compute reversed binary (2-adic) order for -3,-2,-1,0,1,2,3,4,5,6,7,8 maybe. 2-adic expansions (least significant to infinite): 0: 0 0 0 0 ... 1: 1 0 0 0 ... 2: 0 1 0 0 ... 3: 1 1 0 0 ... 4: 0 0 1 0 ... 5: 1 0 1 0 ... 6: 0 1 1 0 ... 7: 1 1 1 0 ... 8: 0 0 0 1 ... -1: 1 1 1 1 ... -2: 0 1 1 1 ... (since -2 = ...1110, LSB 0 then 1s) -3: 1 1 1 1? Wait -3 two's complement 4-bit 1101, 8-bit 11111101: LSB 1, next 0, then 1s. Let's do properly: n mod 2^m. -1 mod: 1,3,7,15 -> bits LSB:1,1,1,1... -2: mod 2=0,4=2(10),8=6(110),16=14(1110): bits 0,1,1,1... -3: mod2=1,4=1? -3 mod4=1 (01) bit1? LSB sequence: mod2=1, mod4=1 (bits 1,0), mod8=5 (101) bits 1,0,1, mod16=13 (1101) bits 1,0,1,1 => 1,0,1,1,1... -4: 0,0,4,12,28: bits 0,0,0,1,1,1... Nonnegative as above. Now order by real 0.b0 b1 b2 ... (binary fraction). Let's list values and reversed prefix: 0: 0.0000... 1: 0.1000... 2: 0.0100... 3: 0.1100... 4: 0.0010... 5: 0.1010... 6: 0.0110... 7: 0.1110... 8: 0.0001... -1: 0.1111... -2: 0.0111... -3: 0.1011... (bits 1,0,1,1...) -4: 0.000111... -5: bits? -5 mod2=1, mod4=3 (11) bits1,1, mod8=3? -5 mod8=3 (011) bits1,1,1? Let's not. Order by real: start with b0=0 block (even 2-adic numbers) then b0=1 block. Within b0=0, b1=0 subblock, etc. This is lexicographic, which for infinite binary sequences with tails 0 or 1 may have order type? Let's sort a finite set by first few bits. For numbers 0..7, reversed bits (b0,b1,b2): 0:000 4:001 2:010 6:011 1:100 5:101 3:110 7:111 Order: 0,4,2,6,1,5,3,7 — exactly bit-reversal! Good. For 0..15, order is 0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15. This is the infinite limit on nonnegative integers? As m→∞, for fixed n, its reversed bits have finite 1s then 0s, so the order among nonnegative integers is: sort by reversed binary fraction. This order is not type ω? Let's see the sequence of nonnegative integers in this order: 0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15,16,24,8? Wait for 0..31, 0,16,8,24,4,20,12,28,2,18,10,26,6,22,14,30,1,... It is not a permutation of N? It includes each nonnegative once, yes as m grows? For fixed n, its position in the order among all N? The order is by real number 0.b0b1... . The set of reversed sequences for N is {binary fractions with finitely many 1s then all 0? Actually nonnegative n has 2-adic expansion with finite 1s and tail 0, reversed has finite 1s then all 0? Yes eventually 0. These are dyadic rationals with finite binary expansion, dense? The set of finite binary sequences (eventually 0) is countable and dense in [0,1]? The order type of finite binary sequences (eventually 0) by lex order is not ω; it's dense? Let's check: sequences eventually 0 include 0.000..., 0.100..., 0.010..., etc. Between 0.000... and 0.100... there are 0.010..., 0.001..., etc. No immediate successor of 0: the least positive? There is no least sequence >0 because 0.000...1 (at arbitrarily far position) is smaller than any fixed positive finite? For n powers of 2, reversed bits 0...01 with 1 at position m: as m→∞, these approach 0. So 0 has no immediate successor. Thus the order on N by reversed 2-adic bits is dense at 0, not type ω. Including negative integers (eventually 1) may fill gaps but still maybe no adjacent pairs? Let's see. The set of all 2-adic integer reversed sequences includes sequences eventually 0 (N) and eventually 1 (negative). Is this set order-dense? Between any two, can we find another eventually 0/1? Likely yes, so not discrete. Thus not a permutation of Z. But maybe we can "coarsen" or "shift" to get a discrete order by taking the 2-adic order and then interleaving? No, if the 3-free property forces the 2-adic order (non-discrete), then no 3-free permutation. However maybe there is a different discrete 3-free order not given by this lex order. Let's check if the sign equations force the order to be the 2-adic lex order. For any 3-free order, for each d, the relation between n and n+d alternates. This is similar to a "2-adic" orientation. There might be many solutions, but the transitive closure might be the 2-adic order. If so, no discrete solution. Let's attempt to prove: The conditions for all d imply that for x x>y in bit-rev? Order 2 before1, yes. x=2 (010), y=3(011): least diff bit0:0 vs1 => 2<3. x=1(001), y=4(100): least diff bit0:1 vs0 => 1>4? Order 4 before1, yes. So finite order: x <_π y iff the first (least significant) bit where x,y differ has x=0,y=1. This relation can be defined on all Z using 2-adic expansions: x <_π y iff at the least significant bit where their 2-adic expansions differ, x has 0 and y has 1. This is the 2-adic lex order. It is translation? It gives a total order on Z_2, dense as argued. The 3-free property for finite intervals exactly matches. Is it the unique 3-free order on Z/2^m? Maybe. If any 3-free order of type Z existed, its restriction to residues mod 2^m might not be bit-reversal due to carries/cross-boundary, but perhaps the sign conditions force this local rule for all pairs at distance power of 2? Let's examine. For finite [0,2^m), 3-free condition for all APs inside may not force the full order, only signs for AP distances. But many 3-free permutations exist for arbitrary n, not necessarily bit-reversal? For n=3, permutations avoiding monotone 3-AP? Only one triple; many. For n=4, maybe more than bit-reversal. So not unique. But for infinite and all APs, maybe more constrained. Let's try to find an explicit 3-free permutation of Z by using a known construction. Search memory: "There exists a 3-free permutation of the integers, for example the sequence a(n) = (-1)^n 2^{v_2(n)} n"? Let's test variants. Maybe define q(n) = (-1)^n (n + 2^{v_2(n)})? Let's generate small and check? Hard. Let's use the sign pattern from 2-adic order but assign discrete ranks by a monotone enumeration of the 2-adic order? Wait if we take the 2-adic order (dense) on Z and then choose a permutation q of Z that enumerates it in a discrete way, the order of values by enumeration is not the 2-adic order, so 3-free property lost. But maybe we can assign integer ranks to each element of the 2-adic order such that the order is not the rank order but ranks chosen to preserve 3-free? No, the monotone condition uses rank order (usual integer order of q(n)). The order must be the rank order. Could there be a 3-free order of type Z that is not the 2-adic order? Let's attempt to build one recursively with a "cut" that is discrete. We need for every point a, the partition of symmetric pairs ±d around a is same side. This condition for all a is equivalent to 3-free. Let's see if a discrete order can satisfy this. Consider the immediate successor relation s(a). For each a, s(a) and its reflection 2a-s(a) are on same side (above) and s(a) < 2a-s(a). This gives a directed edge a -> s(a). The reflection condition implies the order is "spherically symmetric" around each point. Maybe this forces the order to be the 2-adic one, which is dense. Let's try prove no adjacent pairs by showing if a and b are adjacent then contradiction using APs with midpoint not in Z? Need same parity. Suppose a < b adjacent in order. If b-a is even, let m=(a+b)/2. Since a,b adjacent, m is either b. If m_order b. Suppose m < a. Then order m < a < b adjacent. Consider AP (m, a, 2a-m)? The midpoint of m and 2a-m is a. We know m2a-m (middle max), so 2a-m < a. Thus 2a-m < m or between m,a? It is below a. But 2a-m = a+d = b? Wait m=(a+b)/2, 2a-m = (3a-b)/2, not b. Let's compute with variables: Let a=0, b=2d (adjacent in order), m=d. If m 0 between, bad. Thus m cannot be below 0 if 2d above? Let's use symmetry: For point 0, d and -d same side. If d (m) below 0, then -d below 0. But b=2d is above 0 (adjacent to 0). AP (0,d,2d): middle d below 0, endpoints 0 above, 2d above. Sequence 0,d,2d with d below both -> middle min, OK. AP (-d,0,d): endpoints -d below, d below, 0 max, OK. So m below possible. Now a=0,b=2d adjacent, with d<0<2d. Since 2d is immediate successor of 0, no element between 0 and 2d. But -d? Symmetry: d and -d same side (below). -d could be below d or between d and 0. If -d between d and 0, fine. Consider AP (d,2d,3d): middle 2d. We know 2d >0>d. Need 2d extreme relative d,3d. Since 2d max among known, need 3d < 2d. So 3d below 2d. Could be above 0 or below. AP (2d,3d,4d): middle 3d. 3d<2d. To avoid decreasing 2d>3d>4d, need 3d>4d (middle min). Thus even sequence 2d,4d,6d? We already know 2d above 0, 4d? AP (0,2d,4d): middle 2d max => 4d<2d. Could be above 0 but less than 2d. But 2d is immediate successor of 0, so no element above 0 less than 2d. Therefore 4d cannot be above 0; it must be below 0. Thus 4d <0. Then AP (2d,4d,6d): 4d<0<2d. For triple (2d,4d,6d), middle 4d. We know 4d<2d. To avoid decreasing? Actually sequence values in order: 4d < 2d. Need 4d extreme relative 2d and 6d. If 4d is min, need 4d<6d; if max, need 4d>6d (impossible since 4d<2d but could 6d even lower). We can determine from AP (4d,6d,8d) maybe. But also AP (0,4d,8d): middle 4d<0, endpoint 0, so 4d min => 8d>4d. Not enough. Can we derive that some element must lie between 0 and 2d? Maybe -d? AP (d, -d, -3d)? Not. This is getting deep. Let's search memory for exact answer. The phrase "monotone k-term arithmetic progression" in permutations. There is a theorem by J. H. van Lint? "Every permutation of the integers contains a monotone 3-term arithmetic progression" maybe due to? And "there are permutations with no monotone 4-term arithmetic progression" using the 2-adic ordering? Wait if 2-adic ordering is dense, maybe it gives no monotone 4? Hmm. Actually, the 2-adic order (dense) might have no monotone 3-AP at all, but not a permutation. To get a permutation, one can take a discrete enumeration of a 3-free order? No. Maybe the largest forced k is 3, and the construction avoiding 4 uses the 2-adic order approximated by a permutation? Let's explore. If every permutation of Z has monotone 3-AP, answer at least 3. Can we construct a permutation with no monotone 4-AP? Finite 3-free permutations can avoid 3, but if 3 unavoidable, we need avoid 4. There may be a known construction using bit-reversal-like blocks that has no monotone 4-AP but necessarily 3. For example, the 2-adic dense order has no 3; a discrete approximation might have 3 but no 4. Let's test finite bit-reversal sequences: they have no 3, so also no 4. If we can extend finite bit-reversal to a bi-infinite permutation by some method that may introduce 3 but avoids 4, answer 3. How to extend? Maybe take a sequence of blocks of length 2^m with bit-reversal, and arrange blocks with large gaps. Cross-block APs might be monotone of length 3? But maybe can ensure no 4. This seems plausible. But the problem asks "largest k such that in any permutation ... must exist". If 3 unavoidable and 4 avoidable, answer 3. Need prove 3 unavoidable. Is that true? Let's try to prove or find counterexample (3-free permutation of Z). I can attempt a computer-like search for a small bi-infinite pattern with period? A periodic 3-free permutation of Z impossible because values finite. But maybe a simple formula. Let's try to construct a 3-free permutation of Z using a "sine" like q(n)=sin(n)? no. Use q(n) = n + (-1)^n f(n) where f grows. For d=1, need zigzag: q(n+1)-q(n) alternates sign. Let's set q(n) = (-1)^n A_n + B_n? Maybe choose q(n) = (-1)^n n for n≥0 and something for n<0 to cover all integers. We saw even subsequence monotone for q(n)=(-1)^n n? Let's test q(n)=(-1)^n n for all n (with q(0)=0): n: -2 ->2, -1->1, 0->0,1->-1,2->2,3->-3. Duplicates (2). Not permutation. But sign pattern zigzags? 2,1,0,-1,2,-3: 2>1>0>-1 decreasing (bad for d=1 triple -2,-1,0). So not. What about q(n)=(-1)^n (n+1) for n? Duplicates. Maybe q(n) = (-1)^n (2n+1) for n≥0, and for negative n use remaining integers. Let's try define a bijection by splitting n into nonnegative and negative and values into parity classes. We need q(2m) and q(2m+1) for m∈Z. Let's choose values such that even subsequence is a 3-free permutation of some subset S_e, odd subsequence S_o. For d even, need S_e and S_o 3-free. For d odd, cross conditions. If we choose S_e and S_o to be the even and odd integers, and use q(2m)=2 a(m), q(2m+1)=2 b(m)+1 with a,b permutations of Z. Cross d=1: values 2a(m),2b(m)+1,2a(m+1). Since parity of value differs in middle, the middle value's parity not directly separate. But if we ensure b(m) is large relative to a(m),a(m+1) (e.g., b(m)>max(a(m),a(m+1)) and also for negative? Maybe use a and b as 3-free sequences with b(m) between? Let's attempt with a(m)=m, b(m)=M_m large. For d=1, to make middle 2b+1 extreme, choose b(m) > a(m),a(m+1) (max) or < both. We can alternate: b(m) high for even m, low for odd m. For d=3, middle b(m+1) relative a(m),a(m+3); need b(m+1) outside interval. This is like a sequence b that is "2-step" extreme relative to a. Could choose a(m)=0? Not permutation. Maybe choose a and b as interleaved copies of a 3-free sequence with a scaling factor: b(m)=L - a(m) finite. For infinite, use b(m) = M(m) - a(m) where M(m) varies to separate blocks. Let's consider using blocks of m. Partition Z into intervals I_j of length 2^j. Within each block, use a finite 3-free permutation scaled to a value interval. Cross-block APs: if an AP has indices in different blocks, the middle index lies in a block of some scale; we can make its value interval far above/below the endpoint blocks to be extreme. This is a common method for constructing permutations with no monotone k-AP of length ≥4 maybe. For no 3, need every cross-block AP extreme, difficult. For no 4, easier. Let's step back and identify the problem's likely difficulty. The question is short, maybe from an advanced math forum. "What is the largest k such that in any permutation of ℤ there must exist a monotone k-term arithmetic progression x1 <⋯ < xk?" It might be a known open problem? Could be asking for a specific value. We need provide final answer with reasoning. We need be confident. Let's search internal memory of results: - "Monotone arithmetic progressions in permutations" by G. H. ? There is a theorem: For every permutation of the positive integers, there is a monotone 3-term arithmetic progression. The largest k for positive integers is 3? But we proved no 3-free permutation of N, so at least 3. Can there be a permutation of N with no monotone 4-AP? Probably yes. Thus for N, largest k=3. For Z, since no 3-free permutation of N but maybe there is of Z, largest could be 2. The problem specifically uses ℤ, maybe to get answer 2, distinguishing from ℕ. Or they use ℤ to avoid boundary and answer 3? Which is more interesting? Let's see if a 3-free permutation of Z exists in literature. I can try to recall a sequence: "The 3-free permutation of the integers is unique and is given by the 2-adic ordering." But if 2-adic ordering is dense, not a permutation. Wait, maybe the 2-adic ordering of the *integers* is actually discrete of type Z if we define it differently (not lex on reversed infinite bits, but by 2-adic valuation of n with a zigzag). Let's revisit. Define order < on Z recursively: - 0 is somewhere. - The set of numbers congruent to 0 mod 2 is one side, 1 mod 2 the other? For order type Z, the two sides of 0 must be one-sided, not two blocks. But maybe the order is: ... < -2^{m} < ... < -1 < 0 < 1 < ...? No. What is the "2-adic order" as a linear order? There is an order called the "2-adic order" on the 2-adic integers that is not dense? Actually the 2-adic integers with the usual topology are compact, totally disconnected; the order by 2-adic valuation of difference might be a tree order, not linear. A linear order extending the 2-adic valuation could be dense? The set of eventually 0/1 binary sequences is countable; is it dense? Let's verify: Sequences eventually 0 or 1. Take two such sequences s middle max OK. 1,2,3: middle2 rank0 min OK. 2,3,4: middle3 rank4, endpoints2 rank0,4 rank3 -> middle max? 4>3,0 yes. 0,2,4: middle2 rank0 min OK. So finite yes. For arbitrary finite F, maybe can use bit-reversal on a bounding interval and restrict? But APs with endpoints in F and middle outside F irrelevant; order of F can be chosen as the order induced by a finite 3-free permutation of the interval. If F⊂[0,N), take a 3-free permutation of [0,N) (exists) and restrict its order to F. APs within F are within interval, so OK. Thus every finite F is orderable. By compactness, there is a total order < on Z (not necessarily type Z) with no monotone 3-AP. Indeed the 2-adic dense order is one. To get type Z, we need a total order isomorphic to Z. The class of total orders of type Z is not closed under limits. But maybe we can choose a compactness argument with the additional condition that the order is discrete: for each n, there is no element between n and s(n), where s(n) is a chosen successor. We can choose successors recursively? Maybe finite satisfiability plus ability to choose immediate successors might yield a Z-order. We can model a permutation of Z as a bijection q:Z->Z. Equivalent to a bi-infinite sequence of domain elements (in order). We can construct it by choosing a path through a tree of finite partial permutations that extend. At each finite stage, we have an order on a finite interval of positions [-N,N] (in the permutation order) assigned distinct domain integers, satisfying 3-free for APs among assigned domain points that are fully assigned? To extend, we need ensure eventual surjectivity. Since domain Z countable, we can list domain integers and force each to appear. The finite constraints are local: if a 3-AP has all three domain points assigned, their order positions must satisfy extreme. When we assign new positions, we can choose new domain integers (unused) to satisfy constraints with existing assigned APs. Is this always possible? We can use a greedy/backtracking argument if finite extensions exist. But the order positions extend in both directions; we can build a bi-infinite sequence by extending to larger position intervals, adding domain integers. At each extension, we need assign domain integers to new positions such that all APs within the new domain set? Wait APs are in domain integers, not positions. If we build by position, when we assign a new domain integer x to a position, it creates potential APs with two already assigned domain integers if they form x as midpoint/endpoints. We need ensure the order (position) of x is extreme relative them. Since we are placing x at a new position (outside the current position interval, at left or right end), its position is automatically extreme relative any two existing positions (which lie in the middle interval)! Ah, this is key. If we build the permutation order as a bi-infinite sequence by adding new elements only at the ends (left or right), then any new domain integer placed at an end is extreme relative to all pairs of already placed elements. Thus it cannot create a monotone 3-AP where it is the middle. But it could be an endpoint of a 3-AP whose middle is already placed; we need check that the already placed middle is extreme relative the new endpoint and another endpoint. Since new endpoint is outside the current position interval, if the other endpoint is inside, the middle (inside) might be between them in position? Example positions: new at left end, other endpoint and middle inside. If new position < other < middle, then sequence new, middle? Need AP order in domain: suppose AP is (x_new, y, z) with y middle in domain. Positions: x_new at left, y,z inside. For y to be extreme, it must be outside interval between x_new and z. Since x_new is left of all inside, y is inside; y will be between x_new and z if z is to the right of y, or y > z? If z is to the left of y, then y is max (outside). So not automatic; we need choose x_new (value) such that for any AP where x_new is endpoint, the existing middle is extreme. This imposes constraints on which domain integer can be placed at the end. But maybe we can choose from unused integers to satisfy. Alternatively, build by domain intervals: assign order positions to a growing domain interval [-N,N], and place the remaining domain integers outside (in value order?) Hmm. The finite bit-reversal order on an interval can be extended by adding a new domain integer at an end of the order such that all new APs are nonmonotone. If we can do this indefinitely and ensure all domain integers used, we get a 3-free permutation of Z. Let's test. Start with finite 3-free order on {0}. Add new domain integer x. Any 3-AP involving x and two existing points: since existing set size small, choose x so that if x is middle, the new position should be extreme; if x is endpoint, existing middle extreme. If we place x at an end, x as middle is automatically extreme? For AP (y,x,z) with y,z existing, if x placed at end, its position outside y,z, so extreme. Good. If x is endpoint (y,z,x) with y middle existing, we need y extreme relative z and x. Since x at end, if z is the other endpoint inside, y will be extreme if y is on the same side of z as? Suppose positions: x at left end, y,z inside with y position maybe between z and x or outside. For y to be extreme, y must not be between x and z. If x left of all, then y is between x and z iff y position < z position. So we need y position > z position (y to the right of z) for every existing AP (y,z,x) where y middle, z endpoint. Thus x can be placed at left end only if for all existing pairs (y,z) with y middle and x=2y-z, we have pos(y) > pos(z). Similarly right end requires pos(y) < pos(z). This is a condition on x. We can perhaps choose x unused satisfying it. This resembles constructing a "3-free" sequence by adding elements at ends. Since the set of forbidden x for left/right may be finite at each stage, and Z infinite, we can choose an unused x. But we also need eventually include all integers; we can use a priority list and choose when possible? Maybe possible because for any finite set, there are infinitely many x that can be added at an end? Let's check. Given a finite 3-free ordered set S. Define L_forbidden = {x∉S: for some y,z∈S with y middle of (x,y,z)? i.e., x=2y-z, and pos(y) < pos(z)}? For left end placement of x, we need for all such x, pos(y)>pos(z). Since for each y,z, x is determined. So only finitely many x are problematic. Thus infinitely many x can be added to left or right without creating violations with existing APs? Also APs where x is middle: x=(y+z)/2 for y,z∈S; finitely many, and placing at end makes them extreme automatically (since x outside), so no problem even if x in S? If x middle, end is extreme, OK. APs with x and one existing? need three points, so only pairs. Thus we can extend a finite 3-free order by adding a new domain integer at either end, provided we avoid finitely many x. Since Z infinite, there are choices. To construct a permutation of Z, we need a bi-infinite sequence where at each finite stage we have assigned a finite set of domain integers to consecutive positions (an interval in the order), and the order on that set is 3-free for APs within the set. We can grow the set to all Z by always adding unused integers. But if we add at ends, the assigned set in domain is not necessarily an interval; that's fine. We need ensure final order positions form Z (bi-infinite). We can start with one element, then repeatedly add elements to the left or right, producing a finite sequence. In the limit, if we add infinitely many to both left and right, we get a bi-infinite sequence. We also need use all domain integers exactly once. Since at each step we can choose any unused integer (avoid finite forbidden), we can enumerate all integers: at step n, choose the smallest unused integer if it is addable; if not, choose some other. But the smallest unused might be permanently forbidden? Need ensure every integer eventually addable. The forbidden sets change as S grows; a particular x might be forbidden at some stage but become addable later? Adding more elements can only add new forbidden conditions? For left end, x is forbidden if there exists y,z in S with x=2y-z and pos(y)pos(z). If there exists a pair with pos(y)pos(z). For a given pair y,z, it forbids x=2y-z from left if pos(y)pos(z). If pos(y)=pos(z) impossible. Thus each ordered pair (y,z) with y≠z forbids one of left/right for x=2y-z. There are finitely many such x (at most |S|(|S|-1)). Also x that is midpoint of two points in S (x=(y+z)/2) is always addable at an end? If x is middle, placing at end makes x extreme, so no forbidden; but if x is also endpoint for some other pair, may be forbidden. In any case, only finitely many x are forbidden from both left and right? An x could be forbidden left by one pair and right by another, making it not addable at all. The union of forbidden sets is finite. Since Z is infinite, there exists x not forbidden at either end (addable). Good. Thus we can extend indefinitely. Therefore for any finite F, we can extend to any finite size containing F. Now use a standard compactness theorem for countable structures with finite satisfiability and surjectivity? We want a bijection q:Z->Z. Equivalently, a total order < on Z of type Z satisfying 3-free. The finite satisfiability with prescribed finite subsets in both domain and codomain? We can model as a partial bijection between domain Z and position Z. For each finite set of domain integers F and finite set of position integers P (an interval), require a bijection from some finite S⊃F to P? Actually to ensure every domain integer gets a position, we can use a compactness argument on the space of linear orders? Let's be precise. We can view a permutation q:Z->Z as a sequence (q(i))_{i∈Z} of distinct integers, all integers. The 3-free property is: for all a,d, q(a+d) is not between q(a),q(a+2d) in the usual integer order (values). This is a condition on the values q(i) (usual order), not on an abstract order. In our finite extension process, we were ordering domain integers by position and assigning ranks? We need assign actual integer values q(i) (permutation values) that are all distinct and form Z. The 3-free condition uses the usual order of these values. In finite extension, we can assign values as ranks (integers) to the domain points in the order. If the order on a finite domain set S is 3-free, we can label the positions with consecutive integers (e.g., -N..N) to make a finite permutation of that interval. But when we extend, the actual integer labels may need to change to keep a global bijection? In a finite order, the relative order of values matters, not their exact integer labels. For the limit to be a permutation of Z, we need assign integer labels consistent with the order and all used. If we have a total order < on Z of type Z, we can label the elements by their order positions (integers) to get a permutation. Thus it's enough to get a total order of type Z, not assign fixed labels at finite stages. The finite extension process on domain sets with abstract order positions (intervals) can be compactified to a total order of type Z if we require that for each N, the finite order on S_N is placed at positions [-N,N] and S_N grows to all Z. But the abstract order on the limit is the intersection of these finite orders; since positions are fixed intervals, the limit order is a linear order on Z of type Z (each domain point gets a unique position in the limit). Finite satisfiability with nested intervals? We need a coherent sequence: for each N, an order on S_N⊃F_N at positions [-N,N], such that for M>N, the order on S_N restricts to S_N? But S_M may not contain S_N? We can choose a nested sequence S_N? To use compactness, we need finite intersection property of sets of partial orders. We can define for each finite F a set of partial permutations on finite position intervals extending F; finite intersection property if every finite collection can be satisfied. The union of a countable chain? We can use the compactness theorem for first-order logic? The language has constants for domain integers, relation <_order? We want a linear order of type Z. Type Z (discrete, no endpoints, each element has immediate successor/predecessor) is not first-order. But we can construct explicitly via a back-and-forth using the extension property. Simpler: Build the order in both directions step by step while ensuring every domain integer is eventually included. At step n, we have a finite ordered set S_n (positions -n..n) containing {-n,...,n} perhaps. We want to extend from S_n to S_{n+1} while adding one new position at left or right and a new domain integer. To include all integers, at step n we could add the smallest positive/negative not yet included if it is addable; if not, add some other. The concern that a domain integer may become permanently forbidden. But maybe the extension property can be strengthened: For any finite S and any x∉S, there is a finite extension of S (by adding some points at ends) after which x becomes addable. If true, we can ensure all x eventually. Is that true? Let's investigate. Given x forbidden left because there is pair y,z∈S with x=2y-z and pos(y)pos(z). If x is forbidden both left and right, it can never be added at an end! Because the relative order of existing y,z remains. Thus the "add at ends only" construction cannot include some x once it is doubly forbidden. But in finite extension from arbitrary F, if F contains a pair that doubly forbids some x outside F, x can never be included in that branch. However we don't need to include that particular x if we only want some finite S⊃F. But for a full permutation, we must include all x, so this is a problem. Can a finite 3-free order S doubly forbid an outside x? Example S={0,1} with order 0<1. For x=2 (2*1-0? pairs): pair y=1,z=0 (x=2) with pos(y)>pos(z) => forbids right? pair y=0,z=1 (x=-1) pos(y)pos(z), so right forbidden. Is left forbidden? need pair y,z with x=2y-z and pos(y)_pos z. Thus if x has two representations x=2y-z with pairs ordered oppositely, it is doubly forbidden. For a given x, representations correspond to y such that z=2y-x. So two y's. In a 3-free order, can both representations exist with opposite order? If x=2y1-z1=2y2-z2. This means the two pairs (y_i,z_i) are APs with endpoint x. If x is outside S, the triples (z_i,y_i,x) or (x,y_i,z_i) are not fully in S, so no 3-free condition yet. It is possible. If a full 3-free order of type Z existed, for any x, in the final order, when x is considered as endpoint for pair (y,z) in S, the relative order of y,z must be such that x can be placed at the end where x lies in the final order. If x is to the left of all S, it needs y>z for all representations; if to the right, yz (middle max). But y is in domain; its position can be anywhere. In our end-extension construction, x at global end forced y outside [x,z]; if x left end, y must be >z (since y< x impossible). In a full order, x not global end, so more possibilities. Thus the end-extension method is too restrictive. So constructing a full 3-free permutation is not trivial. Let's look for a known explicit 3-free permutation of Z. If it exists, the extension issue must be solved by a clever global order, not simply adding at ends. Maybe use the 2-adic dense order and then "discretize" by taking a bijection from Z to the ordered set that preserves betweenness for APs? Wait if the dense order is 3-free, any order-preserving bijection from Z (with usual order) to a discrete subset of the dense order? We need a subset of the dense order of type Z that is still 3-free with respect to the dense order? If we take a cofinal/coinitial discrete subsequence of the dense order that is "cofinal in cuts", APs involving elements not in subset? No, the domain is all integers; we need order them. If we take the dense order on Z and select a type-Z suborder that contains all elements? impossible; the order itself must be type Z. But perhaps we can modify the dense order slightly (choose immediate successors) while preserving 3-free. The dense order has many cuts; to make it discrete, for each cut choose an adjacent pair? But the set is fixed Z; discreteness means for each element, the next element in the order is some other integer. In the dense order, 0 has a cut with elements approaching from above (powers of 2) and below. To make 0 have an immediate successor, we would need all those approaching elements to be moved below 0 or above some chosen successor, which may break 3-free. Maybe impossible. Let's try to prove no 3-free order of type Z using the approaching powers of 2 argument. Suppose < is 3-free on Z. Consider the point 0. For each d>0, d and -d are on same side of 0. Let A = {d>0: d >_π 0}. Then -A also >_π 0; B={d>0: d<_π 0}, -B<_π0. Since 0 has an immediate successor s in the order. s is some integer. If s>0 (usual) or <0; by symmetry if s>0 then -s>0? Wait s is a domain integer, not necessarily positive. Let s = a. Then -a is on same side as a (by AP -a,0,a), so -a >_π 0 as well. Thus 0's immediate successor a implies -a is also above 0. Since a is immediate, a <_π -a (unless a=-a, no). Thus the set above 0 has a least element a, and -a is another element above. Now consider the point 2a. AP (0,a,2a): middle a. Since 0 < a in order, a must be max relative 0,2a, so 2a <_π 0. Thus 2a is below 0. By symmetry, -2a is above 0. Since a is the least above 0, -2a >_π a? Actually -2a above 0, so a < -2a (unless equal). So above 0 we have a < -a and a < -2a (could order between unknown). Below 0 we have 2a. Now consider 4a. AP (0,2a,4a): middle 2a below 0, endpoints 0 above, 4a? Since 2a<0, to avoid increasing? Values: 0 > 2a. If 4a > 2a, then 2a is min (OK); if 4a < 2a, then 0>2a>4a decreasing (bad). Thus 4a >_π 2a. Could be above 0 or below. If 4a >_π 0, then since a is immediate above 0, a < 4a. If 4a <0, then 2a<4a<0. AP (2a,4a,6a): middle 4a. We know 4a>2a. To avoid increasing 2a<4a<6a, need 4a>6a (middle max). Thus 6a<4a. AP (4a,6a,8a): 6a<4a; to avoid decreasing 4a>6a>8a, need 6a>8a. So the sequence 2a,4a,6a,8a,... in order zigzags: 2a < 4a > 6a < 8a > ... (starting with 2a<4a). Also 2a<0< a. The even multiples 2ka zigzag. Now, because a is the immediate successor of 0, no element of the set above 0 is between 0 and a. In particular, if any even multiple 2ka is above 0, it must be > a. If below 0, it is < all above. The zigzag of even multiples has local minima at 2a,6a,10a? Actually pattern: 2a<4a>6a<8a>... So even index (2a,6a,10a,...) are local minima? Let's see 2a<4a, 4a>6a: 6a<4a, 8a>6a: 6a local min. 10a<8a? pattern continues: 8a>10a, 10a<12a. So minima at 2a,6a,10a,... (4m+2)a. Maxima at 4a,8a,12a,... Can we show one of these minima must be the immediate successor of 0? Not necessarily; they could all be below 0 or above a. Consider the set of even multiples below 0. If infinitely many are below 0, since the order below 0 is type ω* (has a greatest element p), the largest (closest to 0) among them is some 2k a. Could that cause contradiction with APs? If 2k a is below 0 and is the greatest below, then any even multiple above 0 is >a. The zigzag may force alternation crossing 0? Let's see. If 2a<0. 4a>2a; could be below 0 but greater than 2a. If 4a<0, then 6a<4a (below). If 6a<4a<0, 8a>6a but could be below or above. The minima 2a,6a,10a,... are decreasing? Pattern: 6a<4a but relation 6a vs 2a? Not directly. AP (2a,4a,6a) gave 6a<4a, no relation to 2a. AP (2a,? 6a?) middle 4a done. AP (0,2a,4a) gave 4a>2a. AP (0,4a,8a): if 4a? middle 4a. If 4a>2a but maybe below 0. Need 4a extreme relative 0,8a. If 4a<0, since 0>4a, to avoid decreasing 0>4a>8a, need 8a>4a. Already from zigzag 8a>6a but not 4a? Wait AP (2a,4a,6a) middle 4a max => 6a<4a. AP (4a,6a,8a) middle 6a min => 8a>6a. No relation 8a vs 4a. But AP (0,4a,8a) if 4a<0: need 8a>4a (middle min). So 8a>4a. Thus if 4a<0, then 2a<4a<8a? But 8a>4a. Pattern 4a max relative 2a,6a; 8a min relative 6a,10a. Could have 8a above 0. Maybe use 2-adic valuation: The sequence of multiples 2^j a? AP (0,2^j a,2^{j+1}a) imposes 2^j a and 2^{j+1}a alternate relative? For j=0: a>0? a above, 2a below. j=1: 2a below, 4a >2a (could above/below). j=2: 4a,8a: middle 4a extreme. If 4a below, 8a>4a; if 4a above, 8a<4a. So signs of 2^j a relative to 0 may follow a pattern. Could all be below 0 except a? Then a is immediate above; 2a,4a,8a,... below 0. Is that consistent? For 2a<0< a. AP (a,2a,3a): middle 2a below, endpoints a above, 3a? Need 2a extreme. Since 2a< a, to avoid decreasing a>2a>3a, need 3a>2a (could below/above). AP (2a,3a,4a): middle 3a. If 3a>2a and 4a<0, to avoid increasing 2a<3a<4a? If 4a<3a, middle 3a max? Actually 2a<3a; if 4a<3a, sequence 2a,3a,4a with 3a max (non) if 3a>4a. If 4a>3a, increasing bad. So 3a>4a. This gets complex. Maybe there is a known theorem: No 3-free permutation of Z? Let's search memory of "3-free permutation". I recall a result: "There are no 3-free permutations of the positive integers, but there is exactly one 3-free permutation of the integers, up to symmetry, namely the 2-adic ordering." Wait exactly one? If it's the 2-adic ordering, it's not a permutation (not discrete). But maybe they use "permutation" in the sense of a linear ordering, not necessarily order type Z? In some contexts, a "permutation of the integers" could mean a total ordering of Z (a bijection from N to Z?) Hmm. Let's search memory: "3-free permutation" might be defined as a permutation a_1,a_2,... of positive integers such that no a_i < a_j < a_k for i0 and approach 0. So 0 has no immediate successor, but as a linear order it could still be order type ω? No, ω has immediate successor of 0. Dense? Are there elements between 0 and 0.000...01 (power 2^k)? Yes 0.000...1 at later position (power 2^{k+1}) is between 0 and 2^k? Let's compare: 2^k reversed has 1 at position k, zeros after. 2^{k+1} has 1 at k+1, zeros after. Binary fractions: 0...0100... (1 at k) vs 0...0010... (1 at k+1). The latter is smaller. So 0 < 2^{k+1} < 2^k. There is no immediate successor of 0. Thus not type ω. But a permutation indexed by N does not require the value order to be type ω? Wait the sequence a_n induces an order on the range by position: n1 < n2 in positions => a_{n1} <_pos a_{n2}. This order is type ω by definition (positions N). The usual integer order of the values is separate. The 3-free condition uses the usual integer order of values, not the position order? Let's clarify: A monotone AP in a sequence a_n usually means positions n_i form an AP and the values a_{n_i} are monotone in the usual value order. Thus the relevant order for monotonicity is the usual integer order on values, while the AP constraint is on positions. We are not requiring the sequence's order to be 3-free in the sense of an order on domain by value? Actually we can invert: A monotone 3-AP in sequence a (positions AP, values monotone) means for some position AP i position, with positions ordered by N or Z, values ordered by usual integers. For one-sided positions N and values Z, no global minimum in positions? Positions have minimum 0, values no minimum. The well-ordering obstruction for N range used positions as values? Let's set variables consistently. We have two equivalent formulations: 1. Sequence a: P -> V, where P (positions) is N or Z with usual order, V=Z values with usual order. A monotone 3-AP means there exist positions p,p+d,p+2d such that values a_p,a_{p+d},a_{p+2d} are monotone in V (inc or dec). 2. Inverse q: V -> P, q(v)=position. A monotone 3-AP in values v,v+d,v+2d with positions monotone in P. This is a 3-free condition on q: for all value APs, q(middle) extreme relative q(endpoints) in P-order. The AP constraint can be on positions (formulation 1) or values (formulation 2)? Wait in formulation 1, positions are AP, values monotone. In formulation 2, values are AP, positions monotone. Are these equivalent under inverse? If positions AP and values monotone, then in q, the values (not necessarily AP) have positions monotone; the values are an AP? No, in formulation 1, values are monotone but not necessarily arithmetic progression. The problem says "monotone k-term arithmetic progression x1<... position is the sequence to study, with domain Z (values) having APs, range positions (permutation order) having monotonicity. If the permutation is indexed by N, range is N (well-ordered). If indexed by Z, range is Z. If AP constraint on values and positions monotone, then a 3-free permutation of Z (no monotone 3-AP in this sense) is a q: Z -> P (positions) with no value-AP where q-values monotone. If P=N (one-sided), the range is well-ordered. We proved no 3-free q: Z -> N? Let's revisit: We proved no bi-infinite sequence P_m (m∈Z) of positive integers (range N) with no monotone 3-AP in the sequence (domain APs, range monotone) using global minimum 1 in range. That is exactly q: domain Z -> range N. So if positions are N (one-sided), no 3-free permutation of Z (in value-AP sense). Thus every one-sided permutation of Z contains a monotone 3-AP. If positions are Z (two-sided), possible? The range is Z, no well-order. If AP constraint on positions and values monotone, then q inverse: domain positions P, range values Z. If P=N (one-sided), range Z no minimum; maybe 3-free exists? If P=Z, range Z. The problem's notation unclear. Let's infer from "permutation of ℤ". In additive combinatorics, a permutation of ℤ is often a bijection π:ℤ→ℤ. They may consider the sequence (π(n))_{n∈ℤ}. A "monotone arithmetic progression" in π might be π(x_1),...,π(x_k) where x_1,...,x_k is an arithmetic progression and π(x_i) is monotone. They would denote x_1<...Z with no monotone 3-AP in the sense of domain APs. We've been analyzing both; it's the same as value-AP sense after inverse (since both domain/range Z). So if a 3-free permutation of Z (domain APs) exists, answer 2. If not, answer 3. If domain is N (one-sided) and range Z, AP constraint on domain (positions), monotonicity on range values. This is a one-sided sequence of all integers with no monotone 3-AP in positions. Does it exist? The range has no minimum, so maybe yes. The finite bit-reversal limit on N? The 2-adic order on nonnegative values? Let's see. We need a sequence a_n (n∈N) of all integers such that for every n,d, a_{n+d} extreme between a_n,a_{n+2d}. This is a sequence indexed by N with values Z. The earlier obstruction for range N does not apply; values Z unbounded both. Can we construct? Maybe easier: Take a 3-free permutation of Z (if exists) indexed by Z and then enumerate it in a one-sided way? The sequence order would be different (type ω), not the 3-free order. But if we just take the 2-adic order on Z (dense) and list it in a sequence of type ω, the AP constraint is on the sequence indices (type ω), not the dense order. We can perhaps use the 2-adic order to assign values to positions N by a monotone enumeration of the dense order? The AP constraint on positions N is independent of the dense order; we need values at AP positions extreme. This is exactly a 3-free sequence indexed by N with range Z. Does the finite bit-reversal construction extend to N? We can define a_n by the 2-adic order of n? For n∈N, let a_n = the n-th integer in the 2-adic order? But the 2-adic order is dense, no n-th (no least after cuts). However we can enumerate it in a sequence that respects some local property? Maybe use a_n = the integer whose reversed 2-adic bit sequence is the n-th in lex order? That's just a_n = n? Hmm. Maybe the problem uses permutation of ℤ as a bijection ℤ→ℤ, so domain two-sided, to allow the 2-adic dense order? No, a bijection ℤ→ℤ can be seen as a sequence indexed by ℤ, not dense. Let's search memory of exact theorem: "Every permutation of the positive integers contains a monotone 3-term arithmetic progression" maybe is false? Let's test with a one-sided sequence of all integers (range Z) avoiding monotone 3-AP. Can we construct a simple one? Try a_n = (-1)^n n? Range not all. But range Z no minimum, so maybe a_n = (-1)^n (some enumeration) works. For d=1, need zigzag. We can choose a_n = (-1)^n b_n with b_n positive enumeration of N/2? For d odd, endpoints opposite parity? If n,d odd => n+2d odd, middle even? Let's set even positions positive, odd negative. For any odd d, endpoints n,n+2d same parity, middle opposite. If even positive, odd negative, middle extreme. For even d, endpoints and middle same parity; need parity subsequences 3-free with values positive/negative one-sided? Even positions positive indexed by N (m≥0) one-sided range N. We proved no one-sided sequence of N indexed by N with no monotone 3-AP? Yes, global minimum 1 obstruction for domain N and range N. Thus parity separation fails for N domain. But if even positions use positive, odd negative, even subsequence domain N (m≥0) range N one-sided; impossible. If use even positions all integers unbounded both? But even positions are countably infinite one-sided; can map to Z (two-sided) bijectively. For odd d, need middle extreme; sign separation could be even values > odd values, but even subset mapped to an upper unbounded set (N) one-sided; impossible for even subsequence? If even subsequence range N one-sided, impossible. If even range Z two-sided, no global sign separation. Maybe a one-sided 3-free sequence of Z exists using a more subtle construction. If it does, then a permutation of Z indexed by N would have no 3, answer 2. If not, answer 3. Let's try to prove existence of a one-sided 3-free sequence of Z. We can use the compactness/end-extension idea but now range values are Z (no minimum), and positions N one-sided. The global minimum obstruction is gone. Finite satisfiability likely yields a one-sided sequence by König's lemma? Since positions are one-sided, we can build inductively: choose a_0,a_1,... such that every finite prefix is extendable and all integers eventually appear. At step n, the condition for new APs where the new position is the right endpoint (a_{n-1}, a_n? etc.) involve existing values; since values Z unbounded both, we can likely choose a new integer to avoid finitely many bad choices. This is promising! Let's formulate. We want a sequence a_0,a_1,... of distinct integers, eventually all Z, such that for all i,d, a_{i+d} is not between a_i,a_{i+2d}. Build greedily. At step n, we have chosen a_0,...,a_{n-1}. To choose a_n, which triples become complete? Those with indices i,i+d,i+2d = n. So i+2d=n. For each divisor-like d such that n even? Actually i=n-2d, i+d=n-d. Both ≥0. For each d with 0<2d≤n, the triple (n-2d, n-d, n) has a_n as endpoint. To avoid monotone, a_{n-d} (existing middle) must be extreme relative a_{n-2d} (existing) and a_n (new). Since a_n is a new value we can choose. Given two existing values L=a_{n-2d}, M=a_{n-d}, we need choose x=a_n such that M is extreme relative L and x. When is M not extreme? M is between L and x. Thus we must avoid x such that L < M < x or x < M < L. That is, avoid x > max(L,M?) Let's derive: If LM. If x M forbidden. If M M if LM. Also need x not equal to existing values. Thus at step n, the forbidden set for x is a union of finitely many rays: (M,∞) or (-∞,M) for various M. The intersection of allowed conditions is an interval (possibly unbounded) excluding finitely many existing values. Is it always nonempty? We need choose an integer x satisfying all inequalities and unused. The allowed set is [A, B] (integer interval) where A=max of lower bounds (from x>M) and B=min of upper bounds (from xM; another M'M_d; if L_d>M_d, require xM_{d1} and x n-d2. The existing sequence up to n-1 is 3-free. Maybe the signs ensure monotone bounds consistent. Let's test small. Start a0 arbitrary 0. Step1: no constraints (n=1, 2d≤1 none). Choose a1=1 maybe. Step2: d=1: triple (0,1,2): L=a0=0,M=a1=1, L x>1. Choose a2=2? But then 0,1,2 monotone! Wait if L=0,M=1, to avoid M between 0 and x, need x<0? Let's recalc: Sequence L,M,x. M is middle position. M is between endpoints L and x if L1 (LM. If x L > x) not between. If LM. L is M. Thus forbidden x>M. Allowed xM. Thus constraint is simply x < M if L M if L>M. Equivalently, x must be on the same side of M as L? Actually if LM, x>M. So x must lie on the same side of M as L (strictly). Nice. Step2 with 0,1: L=0 x<1. Choose a2=-1 (not 2). Sequence 0,1,-1: triple 0,1,-1 non (1 max). Good. Step3: triples with n=3: d=1: (1,2,3): L=a1=1,M=a2=-1, L>M => x > -1. Choose a3=2? Existing 0,1,-1; x=2 allowed. Sequence 0,1,-1,2. Check d? Only d=1 for n=3? Also 2d≤3 => d=1. Triple 1,2,3 values 1,-1,2: middle -1 min, OK. Step4: d=1: (2,3,4): L=-1,M=2, L x<2. d=2: (0,2,4): L=0,M=-1, L>M => x>-1. Allowed -1-1 and <2 => 0,1; both used. No available! Could choose a3 differently? Let's backtrack. At step3, choose a3=0? used? no 0 used. a3 must >-1, unused: choose a3=0? used. choose a3= -? >-1: 0 used, 1 used, 2 available. Only 2? Also negative? >-1 none except 0. So a3 forced 2. Then step4 no allowed integer. Thus the prefix 0,1,-1,2 cannot be extended. Maybe different initial choices allow extension. Try a0=0,a1=-1 (choose opposite). Step2: L=0,M=-1, L>M => x>-1. Choose a2=1? (unused, >-1). Sequence 0,-1,1. Step3: d=1: L=-1,M=1, L x<1. Allowed unused <1: choose -2? or 0 used, -1 used. Choose a3=-2. Step4: d=1: (2,3): L=1,M=-2 => x>-2. d=2: (0,2): L=0,M=1 => x<1. Allowed -2 x>-2. Choose a3=2? Step4: d1 L=-2,M=2 => x<2; d2 L=0,M=-2 => x>-2; allowed -1,0,1? unused: -1,1? choose -1. Sequence 0,1,-2,2,-1. Check extension step5 maybe. Step5 n=5: d=1: (3,4): L=2,M=-1 => x>-1. d=2: (1,3): L=1,M=2 => x<2. Allowed -1-2, unused: -1,2,3,... Choose a3=-1. Step4: d1 L=1,M=-1 => x>-1; d2 L=0,M=-2 => x>-2; allowed x>-1: choose 2. Step5: d1 L=-1,M=2 => x<2; d2 L=1,M=-1 => x>-1; allowed -1 x<3; d2 L=0,M=-2 => x>-2; allowed -1,0,1,2? unused -1,2. Choose a4=-1? Step5: d1 L=3,M=-1 => x>-1; d2 L=1,M=3 => x<3; allowed -1 x<2; d2 L=3,M=-1 => x>-1; d3 L=0,M=3? triple (0,3,6): L=a0=0,M=a3=3 => x<3. Allowed -1 x>2; d2 L=1,M=3 => x<3; allowed x= ? integer between >2 and <3 none. Fail. Choose a3=4? Step4: d1 L=1,M=4 x<4; d2 L=0,M=-2 x>-2; choose a4=2? Step5: d1 L=4,M=2 x>2; d2 L=1,M=4 x<4; allowed 3? unused choose 3. Sequence 0,1,-2,4,2,3. Step6: d1 L=2,M=3 x>3; d2 L=4,M=2 x>2; d3 L=0,M=4 x<4; allowed 3, choose x>a_{n-1}. So x is beyond a_{n-1} away from a_{n-2}. d=2 gives another bound, etc. The intersection of one-sided bounds is an interval (possibly one-sided). To keep it unbounded in the needed direction, the bounds must not conflict. But multiple d may create both a lower and upper bound, yielding a finite interval. Could we avoid conflict by choosing previous values carefully? Perhaps yes, by making all constraints one-sided (no conflict). For a given n, the signs of a_{n-2d} relative a_{n-d} might all point the same way, so allowed x is a ray. If so, choose a very large/small value. Can we maintain that? For n=4, d=1 sign a2 vs a3, d=2 sign a0 vs a2. In our attempts, they conflicted. Maybe choose a3 such that signs don't conflict. For 0,1,-2, at n=4: d=2 L=a0=0,M=a2=-2 => sign L>M requires x>-2 (lower bound). d=1 L=a1=1,M=a3. To avoid upper bound, need sign LM gives x>M (lower), same direction as d=2, no conflict! At step3 after 0,1,-2, constraint from d=1: L=a1=1,M=a2=-2 => x>-2. We chose positive a3=2,3,4 causing conflict. If choose a3=-1 (negative, >-2), then at n=4: d1 L=1,M=-1 => L>M => x>-1 (lower); d2 L=0,M=-2 => x>-2 (lower). Allowed x>-1, unbounded! But we tried a3=-1 and step5 failed? Let's re-evaluate. Sequence: a0=0,a1=1,a2=-2,a3=-1. Check triples: 0,1,2: 0,1,-2 non (1 max). 1,2,3: 1,-2,-1: middle -2 min? 1>-2<-1 non. n=4 choose x>-1, unused. We chose a4=2. Could choose very large, say 100, to keep unbounded. Let's try a4=100. Sequence 0,1,-2,-1,100. Check step5 constraints: d=1: L=a3=-1,M=a4=100, L x<100. d=2: L=a1=1,M=a3=-1, L>M => x>-1. Allowed -1M => x>99. d2: L=-1,M=100, L x<100. d3: L=0,M=-1? triple (0,3,6): a0=0,a3=-1 => L>M => x>-1. Allowed 99a5 => x>a5; if a4 x L x<100 (upper). d3: L=a0=0,M=a3=-1 => L>M => x>-1 (lower). So allowed interval is (max(lower bounds), min(upper bounds)). d2 gives upper 100. d1 gives either lower a5 (if 100>a5) or upper a5 (if 100 L>M => x>2. d2: L=-1,M=100 => L x<100. d3: L=1,M=-1? triple 1,4,7: a1=1,a4=100? Wait n=7,d=3: indices 1,4,7: L=a1=1, M=a4=100? Actually i=n-2d=1, i+d=4, middle a4=100, endpoint a1=1. L=1 x<100. Allowed 2 endpoint first, then first some second 1). The block values are interleaved, not separated by block. Maybe use a "dyadic" block where values in a block are separated by scale according to binary representation of index. Finite bit-reversal does this. For infinite one-sided, we can take the limit of finite bit-reversal sequences by choosing values not fixed but growing. For each m, S_m of length 2^m is 3-free. We want a single infinite sequence such that for every m, the prefix of length 2^m is 3-free (not necessarily predetermined). We can build S_{2^m} inductively by extending S_{2^{m-1}} to double length while preserving old values? Is it always possible? Our greedy showed finite extensions of arbitrary length may fail if values not chosen well, but perhaps there is a way to double using a transformation. Given a 3-free sequence S of length N with distinct values, can we find a 3-free sequence T of length 2N whose first N terms are S? We need append N new terms. At each append step, constraints may bound new values; but with infinitely many integers, maybe possible if S has a "3-free extendable" property. Bit-reversal prefixes are not extendable by keeping prefix? The length 4 bit-reversal 0,2,1,3 cannot be extended to length 8 by bit-reversal, but maybe can be extended with other values? Let's test S=0,2,1,3. Step4 (n=4): constraints d1 L=1,M=3 => x<3; d2 L=0,M=1 => x<1? Wait L=0 x<1. Allowed x<1, unused: -1. Choose a4=-1. Step5: d1 L=3,M=-1 => x>-1; d2 L=2,M=1? triple 1,3,5: L=a1=2,M=a3=1 => L>M => x>1. Allowed x>1: choose 4. Step6: d1 L=-1,M=4 => x<4; d2 L=1,M=-1? triple 2,4,6: L=1,M=-1 => x>-1; d3 L=0,M=2? triple 0,3,6: L=0,M=2 => x<2. Allowed -1 x>-100; d2 L=2,M=1 => x>1; allowed x>1 choose 4. Step6: d1 L=-100,M=4 => x<4; d2 L=1,M=-100 => x>-100; d3 L=0,M=2 => x<2; allowed 1 a2=-2? Wait a(2)=2a(1)=-2. 3: -2a(1)-1=1. 4:2a(2)=-4 5:-2a(2)-1=3 6:2a(3)=2 7:-2a(3)-1=-3 8:2a(4)=-8 9:4? Sequence: 0,-1,-2,1,-4,3,2,-3,-8,7,6,-5,4,-7,8? Let's list 0..15: a0=0 a1=-1 a2=-2 a3=1 a4=-4 a5=3 a6=2 a7=-3 a8=-8 a9=7 a10=6 a11=-5 a12=4 a13=-7 a14=8? a14=2a7=-6? Wait a7=-3, a14=2a7=-6. a15=-2a7-1=5. Let's recompute carefully using formula: a(0)=0. a(1) = -1. a(2)=2a(1)=-2. a(3)=-2a(1)-1=1. a(4)=2a(2)=-4. a(5)=-2a(2)-1=3. a(6)=2a(3)=2. a(7)=-2a(3)-1=-3. a(8)=2a(4)=-8. a(9)=-2a(4)-1=7. a(10)=2a(5)=6. a(11)=-2a(5)-1=-5. a(12)=2a(6)=4. a(13)=-2a(6)-1=-3? duplicate -3. Bad. Formula not injective. Try a(2n)=2a(n), a(2n+1)=2a(n)+1? identity? no. We need a bijection to Z eventually. The formula should partition Z into even/odd values recursively, like a(2n)=2a(n), a(2n+1)=2a(n)+1 gives a(n)=n. We need a different affine map for odd that still partitions but gives zigzag. For finite bit-reversal, a_{m+1}(2n)=a_m(n), a_{m+1}(2n+1)=2^m+a_m(n) (values lower/upper). For infinite with unbounded both, maybe a(2n)=2a(n), a(2n+1)=2a(n)+1? identity. a(2n)=2a(n)-1, a(2n+1)=2a(n)+1? Ranges even? Let's see partition: if a(n) arbitrary, 2a(n)-1 odd, 2a(n)+1 odd? both odd? no. Need two disjoint subsets whose union Z, e.g., evens and odds: a(2n)=2a(n), a(2n+1)=2a(n)+1 (identity) or a(2n)=2a(n)+1, a(2n+1)=2a(n) (swap). Swapping gives a(n) = bit-reversal? If a(0)=0, a(1)=? a(1)=2a(0)=0 duplicate. Need base. Maybe use a(2n)=2a(n), a(2n+1)=2a(n)-1? Ranges even and odd, but odd values can be negative. If a is a permutation of Z? The map on values: even positions get even numbers 2a(n), odd positions get odd numbers 2a(n)-1. If a(n) ranges over Z, both ranges are all even and all odd, partition Z. Good! Define a(0)=0? Then a(1)=2a(0)-1=-1. a(2)=2a(1)=-2. a(3)=2a(1)-1=-3. Sequence 0,-1,-2,-3,... bad. Try a(2n)=2a(n)+1 (odd), a(2n+1)=2a(n) (even). Base a0=0: a1=0 duplicate. Need base a0? If a0=0, a(0)=0, a(1)=0 no. Use a0=-1? Let's solve affine recurrence for a permutation. Let a(2n)=2a(n)+c0, a(2n+1)=2a(n)+c1 with {c0,c1}={0,1} maybe. For injectivity, the binary representation of a(n) determines a index? If c0,c1 are 0,1, then a(n) is essentially n transformed by bit complement? For a(2n)=2a(n), a(2n+1)=2a(n)+1 gives a(n)=n+a(0)? If a0=0 identity. If a0≠0, not bijection? a(n)=n+a0? Check a(0)=c, a(1)=2c+1, not c+1. Let's not. Our earlier candidate a(2n)=2a(n), a(2n+1)=-2a(n)-1 gave ranges even and negative odd? Not partition (odds only negative). Use a(2n)=2a(n), a(2n+1)=2a(n)+1? identity. Need odd range all odd, but sign of odd values should alternate to zigzag. If a(n) itself zigzags, then 2a(n)+1 follows. Identity bad. What about a(2n)=2a(n), a(2n+1)= -2a(n)+1? Ranges even and odd (all odd if a(n) all Z). Check injectivity? For a given value y, if even y=2x => x=y/2; if odd y=-2x+1 => x=(1-y)/2. Partition. This defines a unique inverse, so if we start with a0? It might be a bijection automatically if the recurrence is invertible: Given y, determine n parity and a(n). Starting from a0, the set of values reachable may be all Z if the affine maps generate Z. Let's test with a0=0: a0=0 a1=1 (since -0+1=1) a2=2a1=2 a3=-2*2+1=-3 a4=2a2=4 a5=-2*4+1=-7 a6=2a3=-6 a7=-2*(-3)+1=7 a8=8 a9=-15 a10=-12 a11=13? Sequence: 0,1,2,-3,4,-7,-6,7,8,-15,-12,13,... Check d=1: 0,1,2 monotone (bad). So not 3-free. Try a(2n)=2a(n), a(2n+1)=1-2a(n) but base a0 maybe negative to avoid initial monotone. a0=-1: a0=-1 a1=3 a2=6 a3=-11 a4=12 a5=-23 a6=-22 a7=23 ... d=1: -1,3,6 monotone (bad). Need a1 not between a0,a2. a2=2a1. For triple 0,1,2: a1 between a0 and 2a1? If a0=-1, a1=3, a2=6: -1<3<6 bad. If a1<0, a0? a1=1-2a0. Need a1 extreme relative a0,2a1. Since a2=2a1. If a1>0, a2>a1; need a0>a1 for a1 min? Actually sequence a0,a1,2a1. If a1>0, 2a1>a1; to avoid increasing a0a1 (a1 min? a0> a1? then a0> a1 <2a1 non). So choose a1< a0. With a1=1-2a0, inequality 1-2a0 < a0 => 1<3a0 => a0>1/3, but a1>0 and a0 positive? If a0=1, a1=-1, then triple 1,-1,-2 decreasing bad? a1<0, a2=-2a1>a2, need a0a1. Nonmonotone if a0 max(a0,a2) or < min. Since a2a0. So need a1>a0. But a1<0 and a1>a0 means a0 even smaller. Inequality 1-2a0 > a0 => 1>3a0 => a0<1/3. Also a1<0 => a0>1/2. Impossible. If a1>0, need a1 min: a1a1) => a1 a0>1/3; a1>0 => a0<1/2. Choose a0? integer none (a0 integer, 1/3-3>-6 bad; a0=1:1,3,6 bad. No. Thus simple 2n affine recurrence not work. Let's search for known 3-free sequence formula. Maybe a_n = (-1)^n floor((n+1)/2) with a 2-adic interleaving of magnitudes: a_{2n}=a_n, a_{2n+1}= -a_n? duplicates. Another idea: Use a_n = v_2(n) - v_2(n+1)? Not distinct. Maybe the largest k is not 2 because 3-free permutation doesn't exist. The finite prefix extendability struggles suggest infinite 3-free sequence may be impossible, even for one-sided Z range. But our greedy might just be unskillful. Let's look for a theorem: "There is no 3-free sequence of integers" perhaps? A sequence of distinct integers with no 3-term monotone AP might be impossible by a simple argument? Let's test. Can we prove no infinite sequence of distinct reals (not necessarily permutation) with no 3-term monotone AP? Wait we found finite examples arbitrarily long? If arbitrarily long finite examples exist, infinite may still be impossible due to compactness? For reals with unbounded range, compactness fails. But maybe a topological argument: The set of infinite sequences with no monotone 3-AP is nonempty? We can construct using the 2-adic dense order? Let's try to define an infinite sequence a_n (n∈N) of distinct reals by taking the 2-adic order on N? The sequence a_n = rank of n in the 2-adic order? The 2-adic order on N is dense (no immediate successor), but we can still enumerate it in a sequence of type N? The enumeration order would not be the 2-adic order; however the 3-free condition for a sequence a_n uses the sequence order (indices) and value order (reals). If we set a_n = f(n) where f is a bijection from N to a countable subset of R, the value order is the usual order of f(n). We need for index APs, f(n+d) extreme between f(n),f(n+2d). This is exactly a 3-free permutation of the value set (with value order). The value set can be any countable ordered set of type η (rationals) perhaps. If we choose the value order to be the 2-adic order (dense) and then map it to reals order-isomorphically, and define f(n) = the value whose 2-adic-order rank is n? Wait the index order is N; we can choose f(n) such that the value order <_value restricted to f(N) is the 2-adic order? But the value order is a linear order of type η (dense) if f(N) is dense in R, while the index order is N. The 3-free condition requires that for each index AP, the middle index's value is extreme in the value order. This is a property of the map f from index set N (with APs) to the value ordered set. It does NOT require the value order to be discrete or type N. We can choose the value ordered set to be the 2-adic order (dense) on the index set itself! That is, choose a bijection f: N -> Z? No, values must be integers (discrete order Z) if permutation of Z. But if we allow values in a dense order, the 2-adic order could work. For integer values, the value order is discrete Z, not dense. But we can map the dense 2-adic order on N to Z via an order isomorphism? A countable dense order without endpoints cannot be order-isomorphic to Z (discrete). So no. However, a sequence a_n of integers induces a value order on the used values that is Z (discrete), but the order of indices by value is a permutation of Z (type Z or ω). The 3-free condition is about the relation between index APs and value order. We can think of it as a total order on the index set (by value) that must be a permutation order (type Z/ω). If the 2-adic order (dense) is 3-free, to get a permutation we need a discrete order approximating it. Maybe impossible because discreteness conflicts. Let's test the no 3-free sequence of integers theorem. Suppose a_n distinct integers, no monotone 3-AP in positions. Consider the first three terms? We can always extend finite. Is there a known infinite sequence? I can try to use the 2-adic order on the indices but assign integer values by a "ranking" that is not the 2-adic order but preserves betweenness for APs. Maybe take a_n = the 2-adic valuation of n? Not distinct. Let's search memory: "A sequence of integers is called 3-free if no three terms form a monotone arithmetic progression. Does there exist an infinite 3-free sequence?" I think yes, and a simple example is the sequence of powers of 2 with alternating signs? Check a_n = (-1)^n 2^n: even subsequence 1,4,16 monotone -> bad. What about a_n = (-1)^n 2^{v_2(n)}? duplicates. Maybe use a_n = (-1)^n (n + 2^{v_2(n)}). Let's test small: n=0? v2(0) issue. For n≥1: n=1: -? 1+v2=2 -> -2 2: + (2+1)=3 3: -(3+1)=-4 4:+(4+2)=6 5:-(5+1)=-6 duplicate 6? sign negative -6 vs positive 6 distinct; okay. 6:+(6+1)=7 7:-(7+1)=-8 8:+(8+3)=11 Sequence -2,3,-4,6,-6,7,-8,11,... d=1 zigzag? -2<3>-4<6>-6<7>-8<11 yes. d=2: even positions (1,3,5,...): 3,-4,7,-8: 3>-4<7>-8 zigzag. Odd positions: -2,6,-6,11: -2<6>-6<11 zigzag. d=3: subsequence positions 1,4,7: 3,6,-8 -> 3<6>-8? middle 6 max, OK for first triple; positions 2,5,8: -4,-6,11 -> -4>-6<11 middle min. This candidate might work? Need check all d and duplicates. Formula a(n)=(-1)^n (n + 2^{v_2(n)}) for n≥1. Values: n=1 -2, 2 3, 3 -4, 4 6, 5 -6, 6 7, 7 -8, 8 11, 9 -10? 9+1=10 -> -10, 10 13, 11 -12, 12 14, 13 -14, 14 15, 15 -16, 16 19... Some duplicates? Positive values: for even n, n+2^{v2(n)}. n=4 ->6, n=5 negative -6. n=6 ->7, n=8->11, n=10->13, n=12->14, n=14->15, n=16->19. Negative values: odd n: -(n+2^{v2(n)}) = -(n+1) for odd; even? v2(n)≥1, positive. Negative values are only from odd n? If n odd, negative; if n even, positive. Negative set: -2,-4,-6,-8,-10,-12,... all negative even. Positive set: for even n, n+2^{v2(n)} (even? n even + power of 2 even) gives positive even values: n=2->3? Wait 2+1=3 odd. v2(2)=1 =>2^1=2? I mistakenly used 2^{v2(n)} but v2(2)=1, 2^1=2, so n=2: 4? Let's correct: 2^{v_2(n)} is the largest power of 2 dividing n? For n=2, 2; n=4,4; n=6,2; n=8,8. I used n+1 for odd (v2=0 =>1) okay. Let's recompute: n=1: -(1+1)=-2 n=2: +(2+2)=4 n=3: -(3+1)=-4 (duplicate -4? n=2 positive 4, negative -4 distinct; but n=3 -4, n? positive 4 distinct) n=4: +(4+4)=8 n=5: -(5+1)=-6 n=6: +(6+2)=8 duplicate positive 8 (n=4). Bad. Maybe use n + 2^{v_2(n+1)}? Let's not. Try a_n = (-1)^n (n + 2^{v_2(n+1)}) for n≥0. Compute: n=0: + (0+1)=1 n=1: -(1+2)=-3 n=2: +(2+1)=3 duplicate? positive 3 and negative -3 distinct; ok. n=3: -(3+4)=-7 n=4: +(4+1)=5 n=5: -(5+2)=-7 duplicate -7 (n=3). Bad. Need a distinct formula. Maybe use a_n = (-1)^n f(n) where f is a permutation of positive integers and f has no monotone 3-AP in its even/odd decimations. We proved no f:Z->N 3-free, but for one-sided domain N, f: N->N 3-free impossible? Yes, one-sided domain N range N no 3-free. So sign separation by parity with positive magnitudes impossible. But if even positions positive, odd negative, even subsequence domain N range N one-sided impossible. Thus no simple parity sign. What if magnitudes are all integers (even positions can be positive/negative) and sign pattern for odd d is not absolute. Let's look at the finite bit-reversal sequence of length 16: 0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15. This is 3-free. It has values 0..15. If we could make an infinite sequence by taking a limit where values are not bounded, perhaps use a_n = the integer whose binary representation is the reverse of n's binary representation truncated at variable length? For n finite, define a_n by reversing the binary expansion of n (without leading zeros). Let's test! For n≥1, write n in binary, reverse the bits, interpret as binary. This is a bijection? Not injective: n=1 (1) ->1, n=3 (11)->3, n=7->7; n=2 (10)->2, n=5 (101)->5? Reversal of 101 is 101=5. Not bijection. If we include trailing zeros? The map n -> bit-reversal with fixed length is not well-defined infinite. What about a_n = the 2-adic bit-reversal of n as a 2-adic integer, then mapped to an integer by some enumeration? Not. Let's search memory of "3-free sequence of integers" example: Maybe the sequence is: 0, 2, 1, 4, 3, 6, 5, 8, 7, ...? That's pairs, has monotone 3 (0,1,2? indices 0,1,2 values 0,2,1 non; indices 1,2,3 values 2,1,4 non; but indices 0,2,4 values 0,1,3 monotone). Bad. Maybe: 0, 3, 1, 2, 4, 7, 5, 6, 8, ...? finite 3-free blocks? Could use the "ruler function" to create a sequence with no 3-AP: a_n = (-1)^n * 2^{v_2(n)} * something. Let's systematically design sign pattern for a one-sided sequence a_n (n≥0) with values Z. We need for all n,d, a_{n+d} extreme between a_n,a_{n+2d}. In particular, for d=1, signs alternate. Let's set a_{2m} local minima, a_{2m+1} local maxima (or vice versa). For d=2, even subsequence a_{2m} zigzags; odd subsequence zigzags. For d=4, subsequence a_{4m} zigzags, etc. Thus for every power of 2, the decimation by 2^r is zigzag. This is exactly a "2-adic zigzag" sequence. We can choose values according to a tree: At level r, the sequence along residues mod 2^r has a prescribed alternating pattern. This suggests using a_n = the sum of signs of bits? But values must be distinct and unbounded. If we only require power-of-2 decimations zigzag, maybe a_n = (-1)^{floor(n/2^r)} for all r? Impossible. But we can assign a_n as the 2-adic valuation of n+1/2? Let's think of a total order on N induced by the 2-adic order (dense) and then take a discrete enumeration that respects the zigzag for all d? The 2-adic order has the property that for any AP, middle extreme. If we enumerate the 2-adic order in a sequence of type N that is "cofinal" and such that index APs correspond to value APs? Wait the APs are on indices N, not on the 2-adic order. If we set a_n = rank of n in the 2-adic order? The rank is not an integer order because 2-adic order dense. But we can choose a_n as a real number in the 2-adic order: Let f:N->R be order-preserving? No, f is a bijection from indices N to a subset of R; the value order is the usual real order. We want the value order restricted to {f(n)} to be such that for index APs, f(n+d) extreme. This is equivalent to defining a total order < on N by m < n iff f(m)R distinct, the order < on N is a linear order of type ω (because indices N enumerated in increasing order of f? Wait f is a sequence indexed by N; the value order < is the order of the values, which is some linear order on N. Since values are distinct reals, < is a countable linear order. It is not necessarily type ω; for a sequence indexed by N, the values can be any countable order, e.g., dense. The index order (N) is separate. The 3-free condition is: for index APs, the middle index is extreme in the value order <. Thus we just need a countable linear order < on N (not necessarily type ω) of the same cardinality, with the 3-free property, and then embed < into Z? Wait values must be integers, so the value order < must be order-isomorphic to Z (if all integers used) or to a subset of Z (if not permutation). If the sequence is a permutation of Z, the value set is Z, so < is order-isomorphic to Z (discrete bi-infinite). If the sequence is a permutation of N, < is type ω. But if the sequence is just a permutation of Z indexed by N, value set Z, < is type Z (because the values are all integers in their usual order). The index order is N. Thus we need a linear order < on the set N (domain indices) that is order-isomorphic to Z (value order) and has the 3-free property with respect to the usual AP structure on N. This is different from a permutation of Z (domain Z). The value order < being type Z is possible even though domain is N. We don't need < to be type N. Great! This may allow the 2-adic dense order? No, < must be type Z (discrete), not dense. But we can choose a 3-free order < on N of type Z. Does such exist? The domain N is one-sided (APs only forward), but the value order is two-sided (type Z), avoiding well-order obstruction. This might be possible. The problem "permutation of Z" could be a bijection from N to Z (list all integers), so domain N, value order Z. Then a 3-free permutation would be an order < on N of type Z with no monotone 3-AP in the index APs. This seems plausible and maybe easier: use the 2-adic order on N (dense) and then choose a discrete type-Z suborder? Wait < is a linear order on the index set N; it must be type Z. We need to order the natural numbers (indices) in a type-Z order such that for every index AP, the middle index is extreme in <. This is exactly a 3-free permutation of Z: list the indices in the order of increasing integer value? Let's align: If a:N->Z is a permutation, define q(v)=position (N) of value v. The value order vZ such that the order of a_n by usual integer order is type Z and 3-free. Equivalently, a permutation of Z indexed by N (one-sided) with no monotone 3-AP in indices. I recall that such permutations exist! The example might be the "2-adic ordering" of the natural numbers enumerated in a bi-infinite sequence? Wait the order < on N of type Z can be given by the 2-adic order? The 2-adic order on N (nonnegative integers) we thought dense at 0, not type Z. But maybe the 2-adic order on all integers restricted to N is type Z? We checked nonnegative reversed bits: 0 has no immediate successor (powers of 2 approach). So not type Z. But maybe another 3-free order on N of type Z exists. Let's attempt to construct using a "shifted" 2-adic order: The 2-adic order on Z (all integers) might be dense, but we can choose a type-Z suborder that contains all N? N is a subset of Z; we need order all N. If the full 3-free order on Z is dense, its restriction to N may be dense or type Z? Let's check the 2-adic order on all Z restricted to N: We saw powers of 2 approach 0 from above, so 0 has no immediate successor within N. Not type Z. But maybe a different 3-free order on Z of type Z exists; restriction to N could be type Z? Not sure. Let's search for an explicit type-Z 3-free order on N. We can use a recursive construction with a "center" that moves. Since order type Z has no endpoints, we can build it as a bi-infinite sequence of indices: ... , i_{-1}, i_0, i_1, ... (these are the natural numbers in increasing value order). The sequence of indices (not values) must be a permutation of N (each natural once). The 3-free condition: for any AP in N, the middle index is extreme in this sequence order. This is exactly a "3-free permutation" of N in the sense of values? It is a permutation of N (the sequence of indices) with no monotone 3-AP? Wait now the values are positions in the sequence (integers, type Z), and the APs are on the domain N. This is a sequence b_m (m∈Z) of natural numbers (indices) such that for any AP in N (values?), hmm. Let's set b_m = the natural number (index) whose value rank is m. Then b:Z->N is a bijection (permutation of N indexed by Z). The 3-free condition: for any AP n,n+d,n+2d in N, the rank b^{-1}(n+d) is extreme between ranks b^{-1}(n),b^{-1}(n+2d). This is equivalent to the sequence b_m? Not exactly. It is a permutation of N indexed by Z with no monotone 3-AP in the *values*? If we take the inverse, q(n)=rank of n, q:N->Z. We need q(n+d) extreme. This is a sequence q_n (n∈N) of integers (type Z) with no monotone 3-AP in positions. That's our a_n. So a_n=q_n. Good. Thus we need a one-sided sequence q_n of all integers, no monotone 3-AP. We were trying to construct it. Let's focus on this. If such q exists, then a permutation of Z indexed by N has no 3; largest k=2. If not, answer 3. Can we prove existence using a known "3-free permutation of the natural numbers into the integers"? I think yes: there is a classic construction by A. S. Besicovitch? Let's try to find one. We can use the 2-adic order on the *rational numbers* and map to integers? Another route: Since the value set is Z (order type Z), we can construct the sequence q_n by assigning integer ranks to natural numbers one by one, ensuring that the final order is type Z. The final order type Z means that for every integer rank r, the natural number assigned rank r exists; equivalently, q_n is a bijection to Z. We can build q_n recursively with the property that the set of assigned values has no eventual bound? At step n, we can choose q_n to keep the allowed interval nonempty and ensure surjectivity by a priority method. The earlier dead ends may be avoidable with a global strategy. Let's formulate the constraints for q_n (n≥0) more abstractly. We need for all n,d: q_{n+d} is not between q_n and q_{n+2d}. This is a set of inequalities. We can view it as a partial order on N: for each AP (i,j,k) with j middle, either j is less than both i,k in value order, or greater than both. This is a disjunctive constraint. We need extend to a total order of type Z on N. This is like a 2-coloring of APs: label each AP by + (middle high) or - (middle low). The labels must be consistent with a total order. For d=1, labels alternate: if q_{n+1} high for triple n,n+1,n+2, then q_{n+2} low for n+1,n+2,n+3, etc. So labels for d=1 are determined up to global flip. For d=2, labels on same parity alternate, etc. We can perhaps choose labels consistently using the 2-adic order, and then realize them by a total order of type Z. The sign pattern for the 2-adic order: For AP (i,i+d,i+2d), the middle i+d is high if the 2-adic valuation of d is even? Or depending on i. In finite bit-reversal, for d=1, middle high at odd positions? Sequence 0,4,2,6,1,5,3,7: triple 0,1,2: middle high; 1,2,3: middle low; so high at odd index. For d=2, triple 0,2,4: middle 2 value2, endpoints0,1? middle high? values 0,2,1 high yes at index2 (even? d=2, n=0, middle n+d=2). n=1: 1,3,5 values4,6,5 high at index3 (odd). So high when middle index odd? For d=2, pattern high at odd? triple 2,4,6 values2,1,3 middle low at index4 (even); 3,5,7 values6,5,7 low at index5? Wait 6,5,7 middle 5 low, index5 odd. So not simply. Maybe label σ(n,d)=+ if floor(n/d) odd? For d=1, floor(n/1)=n: high at odd n? middle index n+d? For triple starting n, middle n+d. High if (n+d) odd? For d=1, high if n+1 odd => n even. In bit-rev, triple 0,1,2 high (n=0 even), 1,2,3 low (n=1 odd). So σ(n,1)=+ if n even. For d=2, triple n,n+2,n+4 high if? bit-rev: n=0 high, n=1 high? triple 1,3,5 values4,6,5 high (n=1 odd), n=2 low, n=3 low. Pattern high for n=0,1; low for 2,3: period 4, high if floor(n/2) even? n=0,1 ->0 even; n=2,3 ->1 odd. So σ(n,2)=+ if floor(n/2) even. For d=3 in length8: triples n=0: 0,3,6 values0,6,3 high; n=1:1,4,7 values4,1,7 low. Pattern period 6? n=0 high, n=1 low, maybe σ(n,3)=+ if floor(n/3) even? n=0 high,1 low. For d=4 only n=0? length? For infinite, a natural label: σ(n,d)=+ iff floor(n/d) is even (i.e., the block of length d starting at n is even-numbered). Then for fixed d, σ(n+d,d) = sign? floor((n+d)/d)=floor(n/d)+1, so flips. Good. This label satisfies the local alternation for each d. Is this label realizable by a total order? It says: For AP (n, n+d, n+2d), the middle n+d is high if the block index floor(n/d) is even, low if odd. In terms of quotient-remainder: n = qd + r, 0≤r4, y=4->1, z=7->7; y low. d=3,n=0: x=0,y=3,z=6 values0,6,3; y high. The difference is n=0 vs1. The least significant bit where x and z differ? x=000,z=110; LSB diff bit0: x0,z1 => x xZ given by T(n) = (-1)^n (n+1) for n even/odd? Let's search systematically for a simple bijection with 3-free property using sign pattern σ(n,d)=(-1)^{⌊n/d⌋} or similar. If we can find a total order realizing a simple sign pattern, maybe formula emerges. Suppose we want q(n) = f(n) where f is a bijection to Z and sign(q(n+d)-q(n)) = (-1)^{⌊n/d⌋} (for all d). This is a strong condition. Can we integrate? For d=1, q(n+1)-q(n) alternates. For d=2, sign(q(n+2)-q(n)) depends on floor(n/2) parity: for n=0,1 positive? q(2)>q(0), q(3)>q(1); n=2,3 negative, etc. This would mean q(2)>q(0), q(3)>q(1), q(4)q2q4... (if q1 high). q2>q0 and q2q1 and q3>q2? possible if q3 high. q40 and <10, choose 5. q(3)>10 and >5, choose 15. q(4)<5 and <15, choose -5. q(5)<10? from d=2 n=3: q5q6? choose q5=8. d=2 n=4: q6q6, OK choose -6. This seems possible. The sign pattern may be consistent and values can be chosen as a bijection? This pattern is exactly q(n) =? It resembles q(n)=(-1)^{floor(n/2)} * something? Let's list required signs for d=1: q0q2q4... (even low, odd high). d=2: n=0,1: q2>q0, q3>q1; n=2,3: q4q4,q7>q5; so even subsequence q0,q2,q4,q6,... zigzags starting low-high-low-high? q0 low, q2 high relative q0/q4, q4 low, q6 high. Odd subsequence q1 high, q3 even higher? q3>q1 and q3>q5? n=3? q5q1, so q3 high; q2>q0 high; q4 low; q5? q5q6, so q5 high relative q4,q6, but q50 q(n)=(-1)^{v_2(n)} n? n=1 v0 -> +1, n=2 v1 -> -2, n=3 v0 -> +3, n=4 v2 -> +4? (-1)^2=+4, n=5 +5, n=6 v1 -6, n=7 +7. Sequence 0,1,-2,3,4,5,-6,7. d=1: 0,1,-2 non (1 max); 1,-2,3 non (-2 min); -2,3,4? middle 3 max? -2<3<4? Actually -2<3<4 increasing! triple indices 2,3,4 values -2,3,4 monotone increasing -> bad. Need even subsequence zigzag; q2=-2,q4=4,q6=-6? -2,4,-6 zigzag; but q3=3,q5=5,q7=7 monotone odd subsequence bad. Need odd subsequence zigzag too. So q(n) = (-1)^{v_2(n)} n fails for odd (all +). Use q(n)=(-1)^{s_2(n)} n (Thue-Morse sign)? n=1 -1,2 +2,3 -3,4 +4,5 -5,6 +6,7 -7. d=1: 0,-1,2 non; -1,2,-3 non; 2,-3,4 non; -3,4,-5 non; good? d=2: even subsequence 0,2,4,6 monotone positive bad. Need signs for even subsequence alternate too. Thue-Morse sign on even n: n=2 sign +,4 sign +,6 sign +, so no. Use q(n)=(-1)^{⌊n/2⌋} n? n=0 0,1 1,2 -2,3 -3,4 4,5 5,6 -6,7 -7. d=1: 0,1,-2 non; 1,-2,-3 decreasing (bad). Not. Use q(n)=(-1)^{⌊n/2⌋} (n+1)? 0? n=0 +1, n=1 -2, n=2 -3, n=3 +4, n=4 +5, n=5 -6, n=6 -7, n=7 +8. d=1: 1,-2,-3 decreasing bad. Need a sign function s(n) such that for each d, s? If q(n)=s(n) r(n) with r(n)>0 increasing, sign(q(n+d)-q(n)) is not just s. But if r(n) dominates, sign is s(n+d)-s(n) when signs differ; if signs same, sign depends on r. To make each d-decimation zigzag, we need for each d and n, the sign of q(n+d)-q(n) alternate. A sufficient condition: q(n+d) and q(n) have opposite signs for all d? impossible. Maybe choose q(n) as a 2-adic tree height: q(n) = (-1)^{bit_0(n)} * 2^{v_2(n+1)}? Let's try to derive from finite bit-reversal values. For n=0..7 bit-rev values: 0,4,2,6,1,5,3,7. These are not sign-separated; they are ranks 0..7. If we extend to Z with ranks Z, maybe q(n) = the integer whose binary representation (two's complement) is the bit-reversal of n's 2-adic expansion, but mapped to Z by a Gray code? For n=0..7, q values 0..7. For n=8..15, bit-rev length8? Not nested. But maybe there is an infinite bijection q:Z->Z given by q(n) = the integer whose binary expansion is the bit-reversal of n's 2-adic expansion, with a suitable interpretation that yields all integers and is 3-free. We dismissed because the order by q values (usual integer order) on the domain may be dense? Wait q is a map to Z; the usual integer order of q(n) is discrete. If q(n) is defined by reversing 2-adic bits and interpreting as a 2-adic integer? It may not land in Z. But we can interpret the reversed bits as an ordinary binary integer with a sign convention that makes it a permutation of Z. For example, for each 2-adic integer n, its reversed bit sequence is an infinite binary sequence (not eventually 0 or 1 necessarily). Not all such sequences correspond to integers. We need a bijection between 2-adic sequences (Z_2) and integer binary sequences (finite with sign). There is no canonical order-preserving bijection. Maybe use a "2-adic Gray code" that maps Z to Z and is 3-free. Let's search memory of explicit formula: q(n) = (-1)^n 2^{v_2(n)} (n+1) ? Test small: n=0? v∞. n=1: -1*1*2=-2 n=2: +2*3=6 n=3: -1*4=-4 n=4: +4*5=20 n=5: -1*6=-6 n=6: +2*7=14 n=7: -1*8=-8 Sequence -2,6,-4,20,-6,14,-8,... d=1 zigzag? -2<6>-4<20>-6<14>-8 yes. d=2: even positions -2,-4,20? Wait positions 0? start n=1: even indices? n=1,3,5: -2,-4,-6 monotone decreasing bad. Need parity subsequences zigzag. Try q(n)=(-1)^n 2^{v_2(n+1)} (n+1)? n=0: +1*1? 2^{v2(1)}=1 =>1; n=1: -2*2? v2(2)=1 => -4; n=2: +1*3? v2(3)=0 =>3; n=3: -4*4? v2(4)=2 => -16; n=4:+1*5=5; n=5:-2*6=-12; n=6:+1*7=7; n=7:-8*8=-64. Sequence 1,-4,3,-16,5,-12,7,-64. d=1: 1,-4,3 non; -4,3,-16 non; 3,-16,5 non; -16,5,-12? middle 5 max? 5> -16,-12 yes. d=2: even n 0,2,4: 1,3,5 monotone bad. Need even subsequence zigzag, so values at indices 0,2,4,8,... should alternate high/low. The 2-adic tree suggests value scale depends on highest power of 2 dividing n? For even subsequence, n=2m; q(2m) should be a 3-free sequence in m. If q(2m)=2 q(m) or -2 q(m), scaling preserves 3-free. Thus a recursive formula q(2m)=±2 q(m), q(2m+1)=±(2 q(m)+1) with signs chosen to make d=1 zigzag might work if the ranges partition Z. Let's revisit affine recurrence with signs that may be non-constant. We want a bijection q:Z (or N) to Z. A natural recursive construction: partition domain into evens/odds; partition range into evens/odds. Define q(2m) = 2 q(m) or 2 q(m)+?; q(2m+1)=2 q(m)+? to cover all integers. To make d=1 zigzag, we need q(2m+1) extreme between q(2m) and q(2m+2). If q(2m) and q(2m+2) are both even (or both odd) and q(2m+1) odd, we can choose the odd value to be above both or below both. If the ranges are evens/odds interlaced, an odd value can be above or below depending on magnitude. For all m, we need q(2m+1) > max(q(2m),q(2m+2)) or < min. If we set q(2m+1) = 2 q(m)+1 and q(2m)=2 q(m), q(2m+2)=2 q(m+1), then q(2m+1) is between? For q(m) large positive, 2q(m)+1 < 2q(m+1) if q(m+1) much larger, causing monotone. Need choose scaling so odd value is outside the interval of adjacent even values. If q(m+1) is not too large? We can design q recursively so that q(m+1) is on the opposite side. Suppose we define q(2m)=2 q(m), q(2m+1)=2 q(m)+1 (identity). Then q(2m+1) is between q(2m) and q(2m+2) if q(m)q(m+1), then q(2m) > q(2m+1) > q(2m+2)? decreasing bad. Actually identity q(n)=n has all monotone. If we swap: q(2m)=2q(m)+1, q(2m+1)=2q(m). Then q(2m+1)=2q(m) is between 2q(m)+1 and 2q(m+1)? Depending. Not. What if q(2m)=2 q(m), q(2m+1)= -2 q(m)+1 (ranges even and odd). Then q(2m)=2q(m), q(2m+2)=2q(m+1), q(2m+1)=1-2q(m). For d=1, need odd value outside interval of evens. If q(m) and q(m+1) are both positive, evens positive, odd negative? If q(m) positive, 1-2q(m) negative, below both (if evens positive). If q(m) negative, odd positive, evens negative, above both. Thus if q(m) has the same sign as? We need q(2m) and q(2m+2) to have the same sign as q(m)? Since q(2m)=2q(m), q(2m+2)=2q(m+1). If q(m) and q(m+1) have the same sign, then odd value is opposite sign and extreme. If they have opposite signs, odd may not be extreme. Thus we need the base sequence q(m) to be sign-constant for all adjacent pairs, i.e., all q(m) positive or all negative, impossible for a permutation of Z. But maybe for a one-sided sequence of Z, we can have q(m) positive for all m? No, must use negative values too. Could use two levels: q(2m)=2q(m), q(2m+1)=1-2q(m) works if the sequence q(m) never changes sign across adjacent pairs; i.e., q is either all positive or all negative, not bijection. So fails. Use q(2m)=2q(m)+1, q(2m+1)=2q(m) (odd/even swapped). For d=1, odd? q(2m+1)=2q(m) even, endpoints q(2m)=2q(m)+1 odd, q(2m+2)=2q(m+1)+1 odd. If q(m) sign? Not. Maybe use four affine maps for four residues to get a 2-adic tree with sign changes only at scales where q(m) sign constant? Could partition range into blocks (intervals) rather than parity, so that odd values are outside adjacent even values by using different intervals. For finite bit-reversal, q(2m) and q(2m+1) are in lower/upper halves of the value interval, not just parity. For infinite, we can use value intervals that are unbounded both? Maybe partition Z into two sets A,B that are not separated globally but at each finite stage separated by a moving boundary. A recursive construction on Z with a "center" that shifts could work. Let's attempt to build a 3-free permutation of Z (domain Z) by a recursive "fold" that keeps order type Z. Use a bijection F:Z->Z that is a homeomorphism of the 2-adic integers? Maybe q(n)=n+(-1)^n 2^{v_2(n)}? Let's test with small computational reasoning for a candidate that might be known: q(n) = n + (-1)^n 2^{v_2(n+1)}. Define for n∈Z? Let's use n≥0 first. q(0)=0+1=1? v2(1)=0 =>1 q(1)=1-2=-1 q(2)=2+1=3 q(3)=3-4=-1 dup. No. q(n)=n + (-1)^n (n+1)? n=0 1,1 -1? 1-2=-1,2 4,3 -3? sequence 1,-1,4,-3,5,-5,6,-7,... d=1 zigzag? 1,-1,4 non; -1,4,-3 non; 4,-3,5 non. d=2: even positions 1,4,5 monotone? 1<4<5 bad. q(n)=(-1)^n (n+1) + 2^{v_2(n)}? messy. Let's search memory of "3-free permutation of the integers formula". I vaguely recall a construction using the Stern-Brocot tree: a(2n+1) = 2a(n)+1, a(4n+2) = -2a(n)-2, etc. Maybe four cases to make a bijection and zigzag. Since finite bit-reversal on [0,2^m) can be expressed recursively: For m+1, S_{m+1}(2n) = S_m(n), S_{m+1}(2n+1) = 2^m + S_m(n). This uses values in [0,2^{m+1}). To extend to Z, we can let the offset 2^m be replaced by a bi-infinite "center" that depends on n, e.g., q(2n) = 2 q(n), q(2n+1) = 2 q(n) + 1? That's offset 1 not large enough. If we use q(2n) = 2 q(n), q(2n+1) = 2 q(n) + 2^{v_2(n)+1}? The offset should make odd values above all even values in the local block corresponding to n. In finite construction, the offset 2^m is larger than the range of S_m, so all odd-index values are above all even-index values. For infinite, we can't have a global offset larger than all values, but we can have a local offset depending on n that dominates the neighboring even values. For d=1, we need q(2n+1) above both q(2n) and q(2n+2). If q(2n)=2q(n), q(2n+2)=2q(n+1), we could set q(2n+1)=2q(n)+C_n where C_n is large compared to |q(n+1)-q(n)| and same sign? But q(2n+1) must be a distinct integer in the remaining range. If we choose C_n = M_n large alternating positive/negative? Maybe. For d=3, q(2n+3) (odd) must be extreme relative q(2n), q(2n+6). This involves q(n+1) offset vs q(n),q(n+3). We need offsets for odd indices to dominate differences in q over larger intervals. This suggests choosing offsets that grow rapidly with the 2-adic scale, like C_n = 2^{v_2(n)+1} * something. The finite construction uses offset 2^m for all n in a block of length 2^m. For infinite, for n with high 2-adic valuation? The block length for which n is in a certain parity class grows as 2^m. We can set the offset for index 2n+1 to be ±2^{v_2(n)+1} times a base. This is reminiscent of the formula q(n) = n + (-1)^{?} 2^{v_2(n)+1}. Let's try to derive a recursive formula for a one-sided sequence q_n (n≥0) that generalizes finite bit-reversal with unbounded values. For finite length 2^m, values are in [0,2^m). The recursion can be written in terms of n's binary length. For infinite, we can define q_n by its binary representation: if n = 2^m + r with 0≤r<2^m? Not. Maybe use the following: For n≥1, let v = v_2(n) (highest power of 2 dividing n). Define q(n) = (-1)^{n/2^v?} 2^v * something. The offset for odd child of m is 2^{v_2(m)+1}. Let's test a candidate: q(0)=0. For n≥1, let m = n/2^v where v=v_2(n), m odd? Maybe define q(n) = (-1)^m 2^v * (m+1)? For n=1 (v=0,m=1): -2 n=2 (v=1,m=1): -? (-1)^1 2*2? = -4 n=3 (v=0,m=3): -? (-1)^3 1*4=-4 duplicate. Maybe q(n) = (-1)^{⌊n/2^v⌋} 2^v (⌊n/2^v⌋+1) duplicates. Let's use a tree: Each node m has two children 2m,2m+1. We assign values to nodes. We want the value interval of the subtree rooted at m to be separated from its sibling? In finite bit-reversal, the value interval of subtree of even child is lower, odd child upper, and their sizes double. For infinite, we can assign to each node m a value interval I_m (an interval of integers, possibly unbounded) such that for each m, I_{2m} and I_{2m+1} are separated (one above the other) and their union is I_m, and the intervals form a nested tree covering Z. Then recursively order values within I_m by the same 3-free pattern. If this tree of intervals can be realized in Z with all intervals nonempty and covering each integer exactly once, and with the order within each interval being a permutation of that interval, then we get a 3-free sequence. For finite, I_m are finite intervals. For infinite, we need a binary tree of intervals of integers, each node's interval is union of two separated subintervals. But in Z, an interval (finite or infinite) can be split into two separated subintervals only if one is finite? Let's see. If I_m is an interval of integers (consecutive). To split into I_{2m} and I_{2m+1} that are separated (all one < all other) and cover I_m, one must be an initial segment and the other final segment of I_m. If I_m is finite, both finite. If I_m is infinite (e.g., all Z or ray), one subinterval will be one-sided, the other one-sided or finite. This can work if the root interval is Z: split into lower and upper halves? But in Z there is no midpoint; choose a cut c: I_{2m}=(-∞,c], I_{2m+1}=[c+1,∞) (one-sided). Recursively, one side one-sided, the other one-sided. This yields a binary tree of one-sided/finite intervals. Could it cover each integer once and yield a sequence? The finite bit-reversal on Z might correspond to a tree where root is Z, left child (-∞,0], right child [1,∞), then split each one-sided interval at its endpoint? But one-sided intervals can be split into a finite/one-sided and a one-sided. This is possible. The order within each interval (as a sequence) would be a permutation of that interval, but the whole sequence indexed by N? If we use intervals as value ranges for blocks of indices, perhaps. Let's try this interval tree for a one-sided sequence q_n. Root interval Z. Split at 0: even indices values ≤0? odd indices values >0? For d=1, if odd values > even values, middle (odd) max for triple even,odd,even? But indices 0,1,2: if 0 even low,1 odd high,2 even low, good. For d=3: indices 0,3,6: 0 even low,3 odd high,6 even low, good. So all odd index values > all even index values handles all odd d. Then even subsequence (indices 2m) values ≤0 must be 3-free; odd subsequence values >0 must be 3-free. But even subsequence domain N (m≥0) range (-∞,0] one-sided (type ω*); odd subsequence range (0,∞) one-sided (type ω). We earlier thought one-sided range with one-sided domain may be impossible for type ω (positive) but possible for type ω*? Let's check: odd subsequence range positive N, domain N. No 3-free sequence of positive integers indexed by N (global minimum 1 obstruction). Thus this split fails. If we split root into even range positive, odd range nonpositive, positive side still fails. Could split into intervals that are two-sided at each level? But separated subintervals of a two-sided interval Z are one-sided. Recursing on one-sided intervals yields one-sided ranges; eventually the positive side impossible. So global parity separation impossible. What if at the next level, the positive side is split into two separated subintervals, one finite and one one-sided. The finite interval can be 3-free; the one-sided positive side still eventually contains a ray to +∞ with a minimum? Any one-sided positive interval has a minimum integer, causing obstruction if the corresponding domain is one-sided N. Thus a branch that is a positive ray with domain N cannot be 3-free. Therefore no binary interval separation tree can work for N domain. For Z domain, each branch domain is Z (two-sided), and a one-sided range type ω* or ω might be possible? We earlier found type ω range with domain Z impossible? Let's re-evaluate: sequence indexed by Z of positive integers (range N) no 3-free due to global minimum 1 and infinite right side. If domain is Z, yes impossible. For range negative (ω*) with domain Z, possible? But a positive ray branch impossible. Thus global separation still impossible for Z domain if one branch positive ray. Hence 3-free permutations may not exist if the AP domain is N or Z and range Z? But finite examples exist; infinite maybe impossible due to well-ordering of one of the unbounded rays. However the order of values in a permutation of Z is type Z, which has both a positive ray and negative ray, but not as convex blocks. The obstruction for a one-sided range applies only if the entire range of a subsequence is one-sided and the domain subsequence has an infinite ray. In a 3-free order of type Z, the positive values (in the value order) are not a subsequence corresponding to a parity class; they are scattered among indices. The even/odd domain subsequences may each use both positive and negative values, so no one-sided range obstruction. This suggests 3-free permutations of Z may exist, with values mixed across parity classes. The interval separation tree failed because it forced one-sided ranges on parity subsequences. A more flexible construction would mix values. Can we use a "dense" 2-adic order and then map it to Z by a non-order-preserving bijection that preserves the 3-free property? The 3-free property only requires betweenness for AP triples, not full order preservation. We could assign integer ranks to the elements of the 2-adic order in a way that each AP triple has middle extreme. Since the 2-adic order already has middle extreme (in its dense order), any rank assignment that is order-preserving on each AP triple would work. We need a global order (ranks) that is a linear extension of the partial order where for each AP triple, the middle is either less than both endpoints or greater than both, according to the 2-adic order. This partial order might have a linear extension of type Z! The dense order is one linear extension; we need a discrete linear extension. The partial order may be finitely satisfiable with type-Z extensions. This is promising. The 2-adic order defines for each AP which side the middle is on (above/below endpoints). We need assign integer ranks consistent with these inequalities and all integers used. This is equivalent to finding a total order < of type Z extending the partial order P_2ad. Does such a linear extension exist? If the partial order is dense (like a tree), maybe yes by a back-and-forth. For example, the partial order "x middle1 max? 3 max OK. 1,2,3 middle2 position0 min? endpoints1 pos3,3 pos1 -> middle min OK. So another. Thus not unique. Infinite conditions may still allow discrete extensions. Let's try to construct a 3-free permutation of Z via a linear extension of a partial order that is "thin". Maybe use a greedy algorithm assigning integer ranks one by one to domain integers, ensuring that for every AP triple, when the third element is assigned, the inequality holds. We can choose the assignment order (not the permutation order) conveniently. For example, assign ranks to domain integers in increasing 2-adic tree order. Since we can choose ranks (integers) freely, we can maintain that the assigned elements form a discrete order (by construction) and extend to all Z. This is similar to building a total order of type Z by adding elements with ranks at the ends? But earlier end-extension had dead ends; however if we assign ranks (values) not necessarily by adding at ends of the current rank interval? To keep the final order type Z, the set of assigned ranks should be an interval in Z at each stage (so that unassigned ranks are outside). Adding a new element with a rank at the end of the current rank interval keeps the assigned rank set an interval. This is exactly the end-extension process in value order (not domain). We found some branches dead. But maybe with a careful choice of domain elements (not values) we can succeed. The domain set assigned also should be finite; we can choose which domain integer to assign next to keep extendable. Since domain is Z, we can assign integers in an order that avoids dead ends, e.g., by 2-adic tree. The final rank interval will be all Z if we extend infinitely in both directions. Let's try a recursive interval assignment: Start with domain 0 assigned rank 0. At each step, assign new domain integers with ranks just outside the current rank interval, in a pattern that makes them local extrema. For a 3-free order, if we assign a domain integer x with a new extreme rank, it is automatically extreme for triples where it is middle. For triples where it is endpoint, we need the existing middle to be extreme; this imposes that x's rank lies on a certain side. If we choose x from a set where this holds, possible. The 2-adic tree can guide: When the current assigned domain set is an interval of residues mod 2^m, the next elements to assign at rank extremes are certain points. Maybe the finite bit-reversal order can be seen as assigning domain integers to ranks in the following order (by increasing rank): For length 8, order (ranks 0..7) domain: 0,4,2,6,1,5,3,7. This sequence of domain elements (by rank) is exactly the 2-adic bit-reversal sequence. It can be generated by starting with 0, then alternately adding elements that are midpoints? For length 4: 0,2,1,3. For length 8: 0,4,2,6,1,5,3,7. This is the order of natural numbers by increasing 2-adic valuation? Let's list by 2-adic valuation of the number? 0 (∞), 4 (v2=2), 2 (v2=1), 6 (v2=1), 1 (0),5(0),3(0),7(0). Not sorted by v (would 0,4,2,6,1,3,5,7). It is sorted by v, and within same v by something. For 8: v2: 4 (2), 2,6 (1), 1,5,3,7 (0). Within v=1, order 2,6 (even lower?); within v=0, 1,5,3,7 (bit-rev). This is a recursive order: first 0, then multiples of 4? Actually sequence: 0, then 4*{0,1}? 4, then 2*{1,3}? 2,6, then odd bit-rev. Hmm. For Z, we can take this sequence and extend to bi-infinite by adding negative numbers in a symmetric way. The rank order (domain elements by increasing value) for a 3-free permutation of Z would be a bi-infinite sequence of domain integers. The finite sequence above is one-sided (starts at rank0). To make type Z, we need prepend negative ranks. Perhaps we can take the bi-infinite 2-adic order sequence (which is dense) and choose a discrete subsequence? But the domain elements in the rank order must be all Z. The finite rank sequences for [0,2^m) can be extended to [-2^m,2^m-1] maybe by a symmetric 3-free permutation. If we have finite 3-free permutations of intervals [-N,N] for N=2^m-1 that are nested, the limit rank sequence might be bi-infinite and perhaps discrete? Let's try to construct finite 3-free permutations of symmetric intervals [-2^m+1,2^m-1] using bit-reversal on the 2-adic residues. For m=1 interval {0}: order 0. m=2 interval {-1,0,1}? Need 3-free? Only AP -1,0,1: 0 extreme. Order maybe -1,0,1? 0 max? -1<0<1? middle 0 between? bad. Order -1,1,0: 0 max? endpoints -1,1 both <0 in order? If order -1<1<0, yes 0 max. So possible. m=3 interval {-3..3}. Can we extend? Maybe use 2-adic order on residues mod 8 mapped to symmetric range. The finite 3-free order on Z/8 (bit-rev) gives a cyclic? We can choose a cut to make a bi-infinite interval. For residues 0..7 order: 0,4,2,6,1,5,3,7. If we map residues to symmetric integers -3..3 via a bijection such that APs modulo 8 correspond to APs in interval? Hard. Maybe use the 2-adic integers and choose a "central" cut in the dense order to get a type-Z order on Z. The dense 2-adic order on Z (reversed 2-adic bits) might have a cut that, if we rotate, yields a discrete order? Dense orders don't have cuts that make discrete by relabeling; the set remains dense in its order topology. But maybe the restriction to the ordinary integers in the 2-adic order is not dense when using a different embedding of Z into Z_2? Z is dense in Z_2, and the 2-adic order restricted to a dense subset of a dense order is dense. So not. Let's look for a proof of nonexistence of 3-free permutation of Z. If we can prove, answer 3 (or 2 depending interpretation). Let's attempt a rigorous proof. Assume q:Z->Z is a permutation with no monotone 3-AP in positions (domain APs). Let < be order by q-value. For every AP a_order a, then 2a-s(a) >_order a. It does not say which is greater. But AP (a,s,2a-s) with middle s: endpoints a and 2a-s. Since s > a, to avoid a < s < 2a-s (if 2a-s > s), we need s > 2a-s (middle max). Thus s > 2a-s. Contradiction if symmetry only gave both above; no contradiction, it determines s > 2a-s. Good! Thus the immediate successor s(a) is greater in order than its reflection 2a-s(a). So we have a < 2a-s(a) < s(a)? Wait s > 2a-s and both > a, so a < 2a-s < s. But s is the immediate successor of a, so no element between a and s. Therefore 2a-s(a) must equal a? Or there is a contradiction! Let's verify carefully. Let a be any integer, s = immediate successor of a in the order (a < s, no element between). Consider the 3-term arithmetic progression in the usual integer order: a, s, 2s-a? The middle is s? Yes, the three numbers are a, s, 2s-a (since s = (a + (2s-a))/2). Are they distinct? If s=a no; distinct. In the order, a < s. For the permutation to be 3-free, s must be extreme relative to a and 2s-a. Since a < s, the only way s is extreme is if s > 2s-a in order (s is maximum); if s < 2s-a, then a < s < 2s-a (monotone increasing in order), which is forbidden. Thus we must have s > 2s-a. Now, what about symmetry around a? For the AP (-? ) 2a-s, a, s: the middle is a, endpoints 2a-s and s. Since a is 3-free, a must be extreme relative to s and 2a-s. We know a < s. For a to be extreme, a must be less than both endpoints (minimum), so 2a-s must also be > a in order. Thus 2a-s is above a. Since s is the immediate successor of a, and 2a-s is above a and distinct from s (unless 2a-s=s => a=s impossible), there must be an element 2a-s between a and s (because a < 2a-s < s? Do we know 2a-s < s? Yes from the AP (a,s,2s-a) we derived s > 2s-a, not 2a-s. Wait the reflection is 2a-s, not 2s-a. I mixed two points. Let's re-evaluate. Symmetry around a uses the AP (2a-s, a, s). The endpoints are 2a-s and s. Since a < s, for a to be extreme, a must be < 2a-s (minimum) OR a > both (impossible since a a. So 2a-s is above a. Is it below s? Not from this. The AP (a,s,2s-a) involves point 2s-a (the reflection of s around s? no, it's the third point continuing from a through s). It forces s > 2s-a (or s< a? no). This tells about 2s-a, not 2a-s. So no immediate contradiction. But we can apply the same to the point 2a-s? It is above a. Since s is immediate successor of a, 2a-s is either > s or maybe equal? It cannot be between a and s because none. Since it's above a and not s, we have s < 2a-s. Thus the order above a starts a < s < 2a-s. Good, no contradiction. This matches earlier: immediate successor a, its reflection -a above and greater than a. Now consider AP (s, 2a-s, 3a-2s)? The middle is 2a-s? Actually s, 2a-s, 3a-2s are in AP with difference a-s. We know a < s < 2a-s. For middle 2a-s to be extreme relative s and 3a-2s. Since s < 2a-s, to avoid increasing s<2a-s<3a-2s, need 2a-s > 3a-2s. So 3a-2s < 2a-s. Could be below a or above. If below a, OK. If above a, since s is immediate successor, it must be 2a-s. Not contradiction. Maybe iterate: The set above a contains s, 2a-s, 3a-2s, ...? Let's define b_0=a, b_1=s, b_{n+1}=2b_n - b_{n-1} (arithmetic progression in usual integers). The order condition along this AP: b_1 is extreme between b_0 and b_2, b_2 extreme between b_1 and b_3, etc. Since b_0 < b_1 in order, b_1 must be max relative b_0,b_2, so b_2 < b_1. For b_2 relative b_1,b_3: we don't know order b_1 vs b_2 (b_2b_2>b_3, need b_3 > b_2 (middle min). Thus the order along this usual AP must zigzag: b_0 < b_1 > b_2 < b_3 > b_4 < ... (starting with b1 high). This is consistent. It doesn't contradict immediate successor unless some b_n falls between b_0 and b_1. But b_1 is immediate, so all b_n (n≥2) must be either > b_1 or < b_0. The zigzag pattern alternates high/low relative neighbors; it could keep all b_n above b_1 or below b_0. Since b_2 < b_1, and cannot be between b_0,b_1, it must be < b_0. Thus b_2 < b_0. Then b_3 > b_2; could be >b_1 or between b_0,b_1 (forbidden) or b_1? Let's check AP (b_1,b_2,b_3) gave b_3>b_2 only. AP (b_0,b_1,b_2) gave b_2b_1, possible. The actual integer values b_n are an arithmetic progression in the domain: b_n = a + n(s-a). Since s is a fixed integer different from a, this is an infinite AP in the domain (if s-a positive or negative). The order along it zigzags: b_0 < b_1 > b_2 < b_3 > ... . This is a one-sided? Actually n∈Z; the AP extends both directions. The order type along this AP is zigzag. Could all b_n for n≥2 be below b_0 or above b_1? The integer progression is unbounded in domain; order values can alternate. No contradiction. But consider the 2-adic multiples: If s is immediate successor of a, then the AP with difference d=s-a. The order along the full arithmetic progression a+nd must zigzag for all n (by d=1 in that AP? The 3-free condition for step d gives zigzag along that AP). Since a and a+d are adjacent in the global order, no other element of this AP can lie between them. Thus for all n≥2, a+nd is either < a or > a+d. The zigzag pattern starting a < a+d > a+2d < a+3d > ... implies a+2d < a (since a+2d (down), a+2d < a+3d (up), a+3d > a+4d (down). Thus even terms a+2d, a+4d,... are local minima; odd terms a+3d,a+5d,... local maxima. We know a+2d < a. Could a+4d be > a+d? The zigzag only says a+3d > a+4d? Actually a+3d > a+4d (down from 3 to4) and a+2d < a+3d. No relation a+4d vs a. It could be below a, between a and a+d (forbidden), or above a+d. If a+4d > a+d, then the local minima a+2dZ such that each orbit under S is a Z-chain. If we take S(n)=n+1 usual, identity order not 3-free. Need a different permutation S of Z with no cycles (one Z-chain) such that the transitive closure (x< S^k(x) for k>0) is 3-free. We can choose S to be a "3-free permutation" in a different sense. Maybe there is a known S: S(n)=2n+1 for n≥0, S(n)=2n for n<0? This is a standard bijection of Z with one Z-chain? Let's test S(n)=2n+1 for n≥0, S(n)=2n for n<0. This maps Z bijectively? Positive: 0->1,1->3,2->5,... nonnegative odd; negative: -1->-2,-2->-4,... negative even. Missing 0? preimage of 0? none (2n+1=0 no; 2n=0 n=0 but negative branch no). Not bijection. The standard bijection from N to Z: S? Not. We need a single orbit under S of type Z. Any fixed-point-free permutation of Z without cycles? Since Z countable, a single Z-chain is a permutation with exactly one orbit of order Z (no finite cycles). Example S(n)=n+1 usual. Other examples via a bijection T:Z->Z: S = T^{-1} (n+1) T, i.e., conjugate of shift. We need S's transitive order 3-free. Could choose T such that S(n) = 2n+? maybe conjugate to shift. For instance, the map S(n)=2n+1 for n≥0, S(n)=2n for n<0 might be conjugate to shift if it is a single chain; need include 0. Standard interleaving: S(2m)=2m+1, S(2m+1)=2m+2, S(-2m-1)=? Let's construct a bijection S with one Z-chain by any permutation T: S = T^{-1} (T(n)+1). The order is T's usual order. T is the permutation (value order) we're trying to find. Circular. Let's search memory one more time. "3-free permutation of the integers" maybe answer: The largest k is 3. There is a theorem by J. Nešetřil? "Every permutation of the integers contains a monotone 3-term arithmetic progression" is equivalent to the nonexistence of 3-free permutations. I recall a result: "There is no 3-free permutation of the integers" maybe proved by a simple parity/adjacency argument. Let's try to prove that. Suppose < is a 3-free order of type Z on the set Z. For each a, the immediate successor s(a) and reflection 2a-s(a) are above a, with s(a) < 2a-s(a) (as we found? Wait we found s < 2a-s because 2a-s above a and not between; yes s is immediate, so s < 2a-s). Thus for every a, we have a < s(a) < 2a-s(a). The map a -> s(a) is the successor. The integer 2a-s(a) is distinct from s(a). Consider the AP with middle 2a-s(a)? The points s(a), 2a-s(a), 3a-2s(a) are in AP. We know s(a) < 2a-s(a). For 2a-s(a) to be extreme relative s(a) and 3a-2s(a), since s<2a-s, it must be max, so 3a-2s(a) < 2a-s(a). It could be below a or between s and 2a-s (impossible because s is immediate above a? Actually interval (s,2a-s) may contain elements; 3a-2s could be there). No contradiction. But note 2a-s(a) is the reflection of s(a) across a. Let r(a)=2a-s(a). We have a < s(a) < r(a). What is the immediate successor of s(a)? Could it be r(a)? If s(a) and r(a) are adjacent? Not necessarily. But maybe AP (a,s(r?),?) forces. Consider the midpoint of s(a) and r(a): (s+r)/2 = (s+2a-s)/2 = a. We already used. The midpoint of a and r(a) is (a+r)/2 = (3a-s)/2, which may not be integer if parity. If s and a have same parity, then r=2a-s same parity, midpoint of a,r is integer m=(3a-s)/2. The AP (a,m,r) has middle m. We know a < r. For m to be extreme, since ar. It cannot be between a,r. If m is between in domain? It is between a and r in usual order. This may impose that m is outside the order interval. Not contradiction. Maybe use parity to force a chain: If s(a) has the same parity as a, then m integer. If different, no. We can choose a such that s(a) same parity? Not guaranteed. Another property: For any a, the sets A_a^+ = {x: a0, a+d in A^+} has a minimum? Not in integer magnitude, but in order. Let s(a)=a+d. Then -d is also in A^+, and s(a) < -d (i.e., a+d < 2a-a-d= a-d? Wait -d domain is a-d? If s(a)=a+d, reflection is a-d. We have a < a+d < a-d in order. Thus a-d is above a+d. Now consider the AP (a-d, a, a+d). Middle a, endpoints a-d and a+d, both above a; a is min, OK. The order segment a < a+d < a-d. What about the AP (a+d, a, a-d) not usual order? The usual sorted AP is a-d, a, a+d if d>0. Good. Now, since a+d < a-d, the interval (a, a-d) in order contains a+d. The point a+2d? AP (a, a+d, a+2d): middle a+d. Since a < a+d, a+2d must be < a+d. Thus a+2d is below a or between a and a+d (impossible) or below a? Since a is immediate? a+2d cannot be between a and a+d because a+d is immediate successor of a? Wait a+d is immediate successor of a, so no element between a and a+d. Thus a+2d is not between; it could be below a or above a+d. But we just need a+2d < a+d, so it cannot be above a+d; it must be below a. Therefore a+2d < a. Great! For d = s(a)-a (assuming s(a)>a in usual? d could be negative; let d=s-a, nonzero. The AP a, s, 2s-a = a+2d. Since a0). Since p0, a+2d is to the right in domain; if d<0, left. In any case, the point obtained by continuing one more step from s in the same domain direction lies below a in order. This is a strong consequence! Let's verify with finite bit-reversal? There are no immediate successors in infinite; for a finite order, take a element with immediate successor in the finite order but at boundary? For finite, the argument that a+2d cannot be between a,s fails if a+2d outside the finite set? In infinite, it's in Z. Good. So for every a, if s(a)=a+d, then a+2d <_order a. Also by symmetry, if p(a)=a-d' (d'=a-p), then a+2d' <_order a? Let's compute: p=a-d' (d'>0). AP (p,a,a+d') middle a. Since p0), AP (a-d, a, a+d). Middle a. p a, so a+2d_a < a < a+d_a. Thus the order along the domain AP a, a+d_a, a+2d_a is (a+2d_a) < a < a+d_a. Middle a+d_a is max, OK. Now, what is the immediate successor of a+d_a? It is some a+d_a+d_{a+d_a}. Could it be a? No, a is predecessor? In a Z order, the immediate predecessor of a+d_a might be a, since a < a+d_a adjacent. Yes, by definition s(a)=a+d_a, so p(a+d_a)=a. Then applying the predecessor constraint to point b=a+d_a: its immediate predecessor is a=b-d_b where d_b=d_a. The AP (a, b, a+d_a) = (b-d_b,b,b+d_b) has middle b. Since a s. The point below a is a+2d. These are different unless d=0. So immediate successor s=a+d gives: - a-d is above s (by symmetry around a). - a+2d is below a (by AP a,s,a+2d). Thus the order around a has ... a+2d < a < s=a+d < a-d ... . What about the AP (a+d, a-d, a-2d)? Usual order if d>0: a-2d, a-d, a? Not. Let's list points in the domain AP with step d: ..., a-2d, a-d, a, a+d, a+2d, ... Order constraints known: a+2d < a < a+d < a-d. What about a-2d? AP (a-d, a, a+d) gave a < a-d? yes a-d > a. AP (a-2d, a-d, a) middle a-d. We know a-d > a. For a-d to be extreme relative a-2d and a, since a-d > a, it must be max, so a-2d < a-d. Could be above/below a. If a-2d > a, then since a-d > a and no immediate? a is immediate? We don't know. If a-2d between a and a-d, possible. But AP (a-2d, a-d, a) only says a-2d < a-d. AP (a-4d, a-2d, a) etc. The order along the full AP ... a+2d < a < a+d < a-d ... may continue zigzag: let's determine signs for step d. For the AP with difference d, the 3-free condition forces the order signs to alternate. We know positions: ... a-2d, a-d, a, a+d, a+2d, ... Order so far: a+2d < a < a+d < a-d. Let's assign order values (ranks) relative: rank(a)=0. s=a+d rank +1 (immediate successor). a+2d rank negative (≤ -1). a-d rank > +1 (≥ +2). Now for triple a-d, a, a+d: middle a, endpoints a-d (rank >0) and a+d (+1). a rank0 is min, OK. Triple a+2d, a+d, a+4d: middle a+d. We know a+2d < a+d. To avoid increasing? Need a+4d < a+d (middle max) or a+4d? Since a+2d < a+d, if a+4d > a+d then a+2d < a+d < a+4d increasing bad. Thus a+4d < a+d. Could be below a or between a and a+d (forbidden because a+d immediate successor of a) or below a. So a+4d < a. Triple a+4d, a+2d? Actually AP a, a+2d, a+4d (step 2d) middle a+2d. We know a < a+2d? Wait a+2d < a, so a+2d is below a. For middle a+2d relative a and a+4d: both a and a+4d? We have a+4d < a. Since a+2d < a, to avoid decreasing a > a+4d > a+2d? Let's apply: endpoints a (rank0) and a+4d (rank negative). Middle a+2d (rank negative). It must be extreme. If a+2d < min(a,a+4d) or > max. Since both endpoints are ≥? a is 0, a+4d negative. If a+2d is less than a+4d, it is min OK. If a+2d is between a+4d and a, bad. We don't know. The zigzag for step d suggested even terms (a+2d,a+4d,?) local minima/maxima: a+2d local min? From step d: a < a+d > a+2d < a+3d > a+4d < ... We have a+2d < a and a+4d < a; relation a+2d vs a+4d? For step d, triple a+2d,a+3d,a+4d: a+3d? Unknown. Triple a, a+2d, a+4d (step 2d) may force alternation in the subsequence a,a+2d,a+4d: since a+2d < a, to avoid decreasing a > a+2d > a+4d, need a+4d > a+2d (middle min). Thus a+2d < a+4d < a. Good! So even subsequence a,a+2d,a+4d,... has ranks: 0 > a+2d < a+4d > a+6d? Wait a+4d < a, and a+2d < a+4d. For triple a+2d,a+4d,a+6d (step 2d), middle a+4d is max? Since a+2d a+6d < ... all below a? Likely. This is a one-sided (below a) zigzag sequence in the domain AP step 2d. It may be possible. No contradiction yet. But the immediate successor property gave a+2d < a. Since a is immediate successor of a? Wait s(a)=a+d immediate, so there is no element between a and a+d. The point a+2d is below a. What about the point a-d? It is above a+d. Thus the order interval between a+2d and a-d contains a,a+d. Maybe the AP (a+2d, a-d, ?) with midpoint? (a+2d + a-d)/2 = (a+d)/2 not integer generally. Can we choose d such that d=1 by applying to all a? If for some a, s(a)=a+1 or a-1, then a+2d = a±2. The point below a might be a+2 or a-2. Not contradiction. Maybe a theorem: In any 3-free order of Z of type Z, the immediate successor of a must satisfy |s(a)-a|>1. Could be. Let's look for known "3-free permutation" existence via online memory. I recall a sequence called "3-free permutation of the positive integers" does not exist, but "3-free permutation of the integers" does exist and is unique: it is the sequence of integers in the order of increasing 2-adic valuation, with a tie-breaking that gives a bi-infinite sequence: ..., -3, 1, -7, 2, -11, 4, -2, 0, -6, 5, -10, ... Let's test this candidate for 3-free? Sequence (in order) of domain integers? Suppose the order (by value) is: ... , -3, 1, -7, 2, -11, 4, -2, 0, -6, 5, -10, ... This looks like pairs (positive increasing, negative decreasing) around 0. Let's test AP -1? The sequence includes 1,2,4,5,8? and negatives -2,-3,-6,-7,-10,-11. It might be the 2-adic order made discrete by taking a specific enumeration of the 2-adic tree: At each scale, list the two children of 0: -1,1? But 0's symmetric pairs must be same side. In this sequence, 0 has neighbors -2 and -6? Around 0: ... -2, 0, -6 ...; -2 and 6? 6 not near. Symmetry around 0 requires if -2 is below 0, then 2 below 0? In sequence, 2 is far to the right of 0? Order left to right: ... -3,1,-7,2,-11,4,-2,0,-6,5,-10... So 2 is left of 0, -2 is left of 0? -2 appears immediately left of 0. Pair ±2: 2 and -2 both left of 0, good. Pair ±1: 1 is left of 0? 1 appears far left; -1 not in list? missing. Pair ±4: 4 left, -4 missing. This sequence not all integers. Maybe the correct 3-free order of Z is: ..., -5, 2, -8, 3, -11, 4, -2, 0, -6, 1, -9, ...? Need symmetry around 0: pairs (±1), (±2), ... must be on same side. If 0's immediate neighbors are 1 and -1? They must be same side, so both above or both below. Suppose order ... < 1 < -1 < 0 < ...? Then pair ±1 below 0, good. The sequence around 0 could be 1,-1,0,-3,3? Let's try to build a discrete 2-adic order by choosing immediate successor of 0 as 1, and reflection -1 also above? If 1 is immediate successor of 0, then -1 above 1 (since reflection). Order: ... < 0 < 1 < -1 < ... . AP 0,1,2: middle 1 max, so 2 below 0. AP -2,-1,0: middle -1? Since 0 < -1? Wait order 0<1<-1, so -1 above 0. AP -2,-1,0 (usual -2,-1,0) middle -1 above 0. Endpoint 0 below -1. For -1 to be extreme, since -1>0, it must be max, so -2 < -1. Could be below 0 or between 0,-1 (impossible because 1 between? 0<1<-1, interval (0,-1) contains 1; -2 could be 1? no). So -2 is either <0 or between 0 and -1? If -2>0, since 1 is immediate successor of 0, -2 cannot be between 0 and1, so -2> -1 or <0? If -2 between 1 and -1 possible. Many. Let's attempt to construct using the immediate successor rule recursively. Choose order type Z sequence (by value) of domain integers. Start with ... , -1, 1, 0, ? Wait we need 0's immediate successor maybe 1, immediate predecessor maybe -2? Let's pick a=0, s(0)=1. Then constraints: - -1 is above 1 (symmetry): order 0 < 1 < -1. - 2 is below 0 (AP 0,1,2): 2 < 0. Now s(1)? Since 0<1 immediate, p(1)=0. For a=1, immediate predecessor 0. AP (0,1,2): middle 1, endpoints 0,2. We have 0<1, 2<0, so 1 max, OK. Immediate successor of 1 unknown. By symmetry around 1, if s(1)=x, then 2-x above x. Also AP (1,x,2x-1) forces 2x-1 < x? and if x above 1, 2x-1 cannot be between 1,x, so below 1. Could choose x=-2? But -2 is domain integer; is -2 above 1? We know 2<0, not -2. Choose s(1)=-2? Then order 1 < -2. Symmetry around 1: 2-(-2)=4 above -2. AP 1,-2,-5: middle -2, since 1<-2, need -5 < -2 (below -2); could be below 1. So order 1< -2 <4. Also 0<1. So segment 0<1< -2 <4. Need check AP ( -2,0,2 ) middle 0? Domain -2,0,2. Order: -2? we have 1< -2, and 2<0<1, so 2 <0 <1< -2. Middle 0 is between 2 and -2? 2<0< -2, so 0 is between -> monotone (2,0,-2 in order? Values: 2 (lowest), 0, -2 (highest); middle 0 between) bad! Thus s(1) cannot be -2 if 2<0. Let's check AP (-2,0,2): middle 0. Endpoints -2 and 2. We have 2<0. For 0 to be extreme, it must be max, so -2<0? But -2>1>0, so 0 is min relative -2? Actually order 2<0<1< -2: 0 is between 2 and -2, bad. Therefore -2 cannot be above 0? But symmetry from s(0)=1 gave -1 above, not -2. AP (0,1,2) gave 2<0. It doesn't force -2. The AP (-2,0,2) then forces 0 extreme: since 2<0, need -2<0 (0 max). Thus -2 must be below 0. Good constraint. So s(1)=-2 would put -2 above 1, contradiction. Thus s(1) cannot be -2. Choose s(1)=4? Then order 1<4. Symmetry: 2*1-4=-2 above 4. AP 1,4,7: middle4 max =>7<4. AP (-2,0,2) requires -2<0. So -2 below 0, consistent. Order: -2<0<1<4<-2? Wait -2 above 4 and below 0 impossible. Let's compute symmetry around 1 with s(1)=4: reflection 2-4=-2 must be above 1 (since s above 1), specifically 1<4< -2? But AP (-2,0,2) requires -2<0. Contradiction. Thus s(1)=4 impossible. In general, if s(1)=x>1, reflection 2-x must be above x. AP (2-x,1, x?) not. But AP (2-x,? ) maybe forces. If x>2, reflection negative. AP (2-x,1,x) is not an AP? (2-x,1,x) differences 1-(2-x)=x-1, x-1 yes AP. Middle 1. We know 0<1< x. For 1 to be extreme, since 0? Wait triple endpoints 2-x and x, middle1. If 1 < x, to avoid 2-x <1< x, need 1 > 2-x (middle max). Thus 2-x <1, true if x>1. No constraint on 2-x relative 0. But AP (-? ) For x=4, reflection -2. AP (-2,0,2) we used (endpoints -2,2, middle0) not directly. We know 2<0 from s(0)=1. If -2>1>0, then -2 and 2 are on opposite sides of 0? 2<0< -2, so 0 between, bad. Thus any x with reflection 2-x and x? The point 2? We know 2<0. For AP (2-x,0,x?) no. We derived: s(0)=1 implies 2<0. Also for any y, AP (y,0,-y) requires y and -y same side of 0. Since 2<0, -2<0. Thus -2 cannot be above 1. Therefore s(1) cannot have reflection -2 above. More generally, s(1)=x implies 2-x above 1, so 2-x >0 in order. By symmetry around 0, if 2-x >0 then x-2 >0? Wait pair ±(x-2): if 2-x >0 then -(2-x)=x-2 >0. In particular, if x>2, then x-2 positive >0. AP (0, (x-2)/? ) Not. We know 2<0, so -2<0. If x=4, reflection -2 must be >1, contradict -2<0. If x=3, reflection -1 must be >1, but symmetry? Pair ±1: s(0)=1 gave 1>0, so -1>0, no contradiction (could -1>1). But AP (1,3,5)? Let's test s(1)=3. Order 0<1<3< -1? symmetry 2-3=-1 above 3. AP 0,1,2: 2<0. AP -1,0,1: endpoints -1,1 both >0, 0 min OK. AP 1,3,5: middle3 max =>5<3. AP -2,-1,0: middle -1 >0, endpoint0< -1, need -2< -1? Could be. AP 0,? -1? OK. AP -1,1,3: middle1? Domain -1,1,3 (difference2). Order: -1? above3? We have 0<1<3< -1. Middle 1 is between -1 and 3? 1<3< -1, so 1 is min? Endpoints -1 (highest),3 (middle), middle 1 lowest -> extreme, OK. AP 1, -1, -3 (difference -2): middle -1, endpoints1,-3. Need know -3. Symmetry around0: 3<0? AP 0,3,6? Since 3>0, s? not immediate. AP 0,3,6: middle3>0 =>6<3. Could be >0 or <0. If 3<0? no. Maybe s(0)=1, s(1)=3 is possible. Let's try build order segment: 2<0<1<3<-1. Need place -2<0 (since 2<0). Also -2? AP -2,0,2: endpoints -2,2; 0 must be max (since 2<0) => -2<0 OK. Place -2 maybe below 2? Let order -2<2<0<1<3<-1. Check AP -2,-1,0: middle -1 >0, endpoint -2<0, for -1 max need -2< -1 OK. AP -3,-2,-1? unknown. Now s(-1)? Since 3<-1 immediate? We set 3 immediate successor of1, not of -1. There may be elements between 3 and -1; we can choose none for now. If -1 is immediate successor of3, then p(-1)=3. For a=-1, immediate predecessor 3? Difference -4. AP (3,-1,-5) middle -1: 3<-1, so -1 max => -5<-1 (below -1). Could be. This is complex but maybe a consistent order exists. Let's see if we can continue to construct a full order with s(0)=1,s(1)=3,s(3)=-1? The order would have a cycle? In a Z order, successor map must be a single chain. If s(0)=1,s(1)=3,s(3)=-1, then predecessor of -1 is3; successor of -1 unknown. This chain can continue: ...,0,1,3,-1,... . Is this a possible Z-chain? We can define successor map S: 0->1,1->3,3->-1,... and extend to all Z as a single chain. Need 3-free constraints. This chain order so far: ... < 2? <0<1<3<-1< ... . Let's test AP 0,3,6: middle3. We have 0<3; need 6<3. So 6 is below 3, could be below0 or between0,3 (impossible if 1 immediate? 0<1<3, interval (0,3) contains1; 6 could be between1,3 or below0). If 6 between1,3, then AP 1,? no. Choose 6 below0. Then by symmetry around0, -6>0 (since 6<0). -6 could be above? Pair ±6 same side: 6<0 => -6>0. Where? Maybe above -1. AP -6,0,6: 0 min? endpoints both? 6<0, -6>0, so 0 between -> bad! Wait symmetry around0 says 6 and -6 same side. If 6<0, then -6<0, not >0. I made sign error: For AP (-6,0,6), middle 0 extreme means endpoints same side. If 6<0, then -6<0. So -6 below 0. Good. So both 6 and -6 below 0. Then AP 0,6,12: middle6<0, endpoint0>6, need 12>6 (middle min) or 12? Could be. This construction may be possible. It suggests 3-free permutations of Z may exist. Let's search for a known explicit chain: S(0)=1, S(1)=3, S(3)=-1, S(-1)=? Maybe S(n)=? The chain 0,1,3,-1,? Looks like adding powers of 2? 0,1,3,7,15? But we have -1. Maybe S(n)=2n+1 for n≥0, S(n)=2n for n<0? Chain starting 0->1->3->7->15... not single Z (missing negative). If we allow negative after some: 0,1,3,-1,-3,-7,...? That's two rays. What if the order is: ..., -3, -1, 1, 3, 7, -7, ...? Not. Let's try to use a computer-like search for a small 3-free order of Z by specifying successor chain. We can choose the order sequence (domain integers) u_{-∞},...,u_0,u_1,... . Let u_0=0. Choose u_1=1 (successor of 0). Constraints forced u_2 (successor of1) cannot be 2? Let's test u_2=2. Order 0<1<2. AP 0,1,2 monotone (bad). So u_2≠2. u_2=-1? Order 0<1<-1. AP 0,1,2: middle1 max => 2<0. AP -1,0,1: middle0 min? endpoints -1,1 both >0? -1? In order -1>1>0, so endpoints -1 and1 both >0, 0 min OK. AP 1,-1,-3: middle -1? order 1< -1, so -1 max => -3< -1. Could be. u_2=-1 possible? We earlier tried s(1)=-2, not -1. Let's set order 2<0<1<-1. Check AP 0,1,2: 0<1,2<0 => 1 max OK. AP -1,0,1: order 0<1<-1, endpoints 1,-1 both >0, 0 min OK. AP 1,-1,-3: need -3 < -1? (since 1<-1, middle -1 max) choose later. AP 0,-1,-2: middle -1? order -1>0, endpoint0< -1, need -2< -1? or -2? AP (-2,-1,0) middle -1: endpoints -2,0. Since -1>0, to avoid 0< -1? Wait order 0<1<-1. If -2<0, then 0< -1, middle -1 max if -2< -1? Actually endpoints -2 and0; middle -1. -1 is above 0. For -1 to be extreme, it must be > both endpoints, so -2< -1. If -2>0, impossible (would be between0,-1 or above -1). Thus -2< -1. But symmetry around0: 2<0 => -2<0, consistent. Need -2< -1 (more above? If -2< -1 and -1 above0, order -2<0<1<-1? -2 could be below0). Good. Now u_3 (successor of -1). We have order 2<0<1<-1. What can u_3 be? It must be an unused integer above -1. Constraints: AP 1,-1,-3 middle -1. Since 1<-1, for -1 max need -3< -1 (below -1), no restriction on u_3 except if u_3 is -3? If u_3=-3, then -3 above -1, contradiction (needs -3< -1). So u_3≠-3. AP 0,-1,-2: middle -1, endpoint0, -2. We need -1 max, so -2< -1. If u_3=-2, then -2 above -1, contradiction. So u_3≠-2. AP -1,1,3: middle1? order 1< -1, so 1 min? endpoints -1,3. Need 1 extreme. Since 1< -1, for 1 min need 3<1? If u_3=3, then 3 above -1, order 1< -1<3; middle1 is min (1< -1<3) OK. So u_3=3 possible. Let's try u_3=3. Order 2<0<1<-1<3. Check AP -1,1,3: middle1 min OK. AP 1,3,5: middle3? order 3 highest so far; need 5<3. So 5 below3 (could be below0 or between). AP 0,3,6: middle3 max =>6<3. AP -2,-1,0: need -2< -1; if -2 not placed, must be below -1 (i.e., less in order than -1). Since -1 is at position before3, -2 could be below0 or between1,-1? Between1,-1 none (adjacent? We have 1<-1 adjacent if no u between; currently yes. So -2 must be <0. Also 2<0, so -2<0. Could be <2 or between2,0. Now u_4 successor of3. Unused candidates. AP 1,3,5: middle3 max =>5<3. If u_4=5, bad (5 above3). AP 3,? etc. Choose u_4=2? 2 unused? Actually 2 already placed below0, so used. Unused: -3,-4,4,5,6,... but 5,6 must be below3, so cannot be successor (above3). -3? AP 1,-1,-3 required -3< -1, but successor above3, bad. -4? AP? -4 with 0,4? symmetry: 4? not placed. AP -4,-2,0: middle -2? not placed. Could choose u_4=4? AP 0,3,6 doesn't involve4. AP 3,4,5: middle4? If u_4=4 above3, then 5 must be below4; possible. But AP 1,3,5 requires 5<3, so 5 below3, cannot be above4? If 4 above3, 5 below3<4, AP 3,4,5 middle4 max? endpoints3,5; 4>3, if 5<3, 4 max OK. So u_4=4 possible? Check AP 0,4,8: middle4? not placed 8. AP -? Let's test u_4=4: order 2<0<1<-1<3<4. Check AP -1,3,7: middle3? order -1<3, endpoint7 unknown. Need 3 extreme: since -1<3, to avoid -1<3<7 need 7<3. So 7 below3. AP 0,2,4: middle2? order 2<0<4? Wait order 2<0<...<4. AP 0,2,4 in usual (0,2,4) middle2. Endpoints0,4; order 2<0<4, so 2 is min? 2<0<4, middle2 extreme (min) OK. Good. AP -2,0,2: middle0. Order -2? must be <0, 2<0. Need 0 max, so -2<0 OK. AP 2,4,6: middle4? order 2<...<4, endpoint6? Need 4 max if 6<4, so 6 below4. Could be below3. OK. This chain 0,1,-1,3,4,... seems possible? Let's continue a bit to see if dead end. Placed order: 2,0,1,-1,3,4 (left to right). Used { -1,0,1,2,3,4 }. Constraints for unplaced: - 5<3 (from AP1,3,5) - 6<3? AP0,3,6: 0<3, need 6<3. Also AP2,4,6: 2<4, need 6<4 (weaker). So 6<3. - 7<3 (AP -1,3,7). - 8? AP0,4,8: 0<4, need 8<4; AP4? maybe 8<4. - 9? AP1,4,7? Wait 1,? 7? AP1,4,7 middle4: 1<4, need 7<4 (consistent 7<3). AP3,4,5 middle4: 3<4, need 5<4 (consistent 5<3). AP2,5,8? not. - -2<0 and < -1? AP -2,-1,0 required -2< -1? We had -1 max, endpoints -2,0; -1>0, so -2< -1. Since -1 is at position before3, -2 must be < -1 (could be <0 or between1,-1? no adjacent, so <0). Thus -2<0. - -3< -1 (AP 1,-1,-3) and symmetry? 3>0 => -3>0? Wait pair ±3: 3>0, so -3>0. But AP 1,-1,-3 required -3< -1? Let's check AP (-3,-1,1) middle -1. Order currently 1<-1<3. Endpoints -3 and1. If -3>3? Then order 1< -1<3< -3: middle -1 between? 1< -1< -3, bad. If -3 between -1 and3? 1< -1< -3<3: middle -1? endpoints -3,1; -1 between? 1< -1< -3, yes between (since -1 >1 and < -3) bad. If -3<1? then -3<1< -1? middle -1 max? endpoints -3,1, -1 above both if -3<1 and -1>1, good. So -3 must be below 1. But symmetry around0: 3>0 implies -3>0 (same side as 3). Thus -3>0. It can be between0 and1? 0<1 adjacent? Currently 0<1 adjacent (no element between). So -3 cannot be between0,1. It could be below0? No symmetry says >0. Contradiction! Therefore u_3=3 with u_4=4? Wait symmetry around0: AP (-3,0,3), middle0. We have 0<1<-1<3<4, so 3>0. For 0 to be extreme, -3 must be >0 as well. Yes. AP (-3,-1,1): middle -1. We have 1<-1<3<4. For -1 to be extreme relative -3 and1. Since -1>1. To be max, need -3< -1. Also -3>0. The interval (0,-1) in order is 0<1<-1, contains 1. If -3>0 and -3< -1, it must be 0<-3< -1. It could be between0 and1 or between1 and -1. But 0 and1 are adjacent (no element between) because order ...0<1...; 1 and -1 are adjacent (no element between) currently. Thus no place for -3. Therefore the current finite order cannot be extended to include -3. This is a dead end, similar to before. Could we have inserted some elements between 0,1 or 1,-1 earlier to accommodate -3? But 0 and1 were chosen adjacent (s(0)=1). If we want -3 between 0 and -1, we need not have 0<1 adjacent or 1<-1 adjacent. But s(0)=1 means 0<1 adjacent; s(1)=-1 means 1<-1 adjacent. Thus no room. Maybe choose s(1) not -1, leaving room. This suggests the immediate successor choices create gaps that force reflections into adjacent intervals, causing dead ends unless gaps are left. To have a full order, we may need s(0) not 1 but a value d such that the interval between 0 and s(0) has enough room for reflections. In a Z order, every interval between adjacent elements has no room. Thus if s(0)=d, then 0 and d adjacent; the reflection -d must be above d, and the point 2d below 0. Also reflections of s(0)? The point -? may need room. To avoid dead ends, perhaps s(0) must be such that the required reflection can be placed above d without forcing a point into the adjacent interval. In our choice d=1, reflection -1 above d, and point 2 below0; then point -3 (reflection of3) needed room between0 and -1, causing dead end. If choose d large, maybe similar with other points. This line of reasoning may prove no discrete 3-free order: Adjacent pair a,s forces some other point to lie in the adjacent interval (a,s) or (s,2a-s) etc. Let's try to formalize. Let a < s adjacent. Let d=s-a. We know a+2d < a and a-d > s (as above). Consider the point t = a - d (reflection of s). We have a < s < t. Consider the AP (t, a, s) used. Now consider the point u = a + 3d? AP (a+d, a+2d, a+3d) maybe. We know a+2d < a. What about a-2d? AP (a-2d, a-d, a) middle a-d. Since a-d > a, for a-d max need a-2d < a-d. It could be above t or below a. If a-2d > t, then order a a > a+2d? Wait endpoints a-2d and a+2d, middle a. If a-2d > a, then order a+2d < a < a-2d, a is between -> bad. Thus a-2d < a. So we get: a+2d < a, a-2d < a. Also a-d > s > a. Now consider AP (a-2d, a-d, a) middle a-d. Endpoints a-2d ( a > a-2d. For a-d to be extreme, it is max (above both), OK. AP (a-3d, a-d, a+d) middle a-d. Endpoints a-3d and a+d=s. We know s < a-d. For a-d to be extreme, since a-d > s, it must be max, so a-3d < a-d (no new). Could be above a-d? no, must be <. AP (a-2d, a, a+2d) gave a-2d a-2d > a-4d? If a-4d < a-2d, middle min OK; if a-4d > a-2d, middle max? Need determine. The subsequence a-4d,a-2d,a (step 2d) must zigzag: since a-2d < a, to avoid decreasing a > a-2d > a-4d, need a-4d > a-2d (middle min). Thus a-2d < a-4d < a? Wait middle a-2d min means a-2d < a and a-2d < a-4d. So a-4d > a-2d, but a-4d could be a. If a-4d > a, then since a-2dZ is a permutation, its values are unbounded above and below. Let M be a value that is a local maximum (there are infinitely many). Choose a local maximum that is globally maximum among local maxima in some half-line? Since integers no max. But maybe use a local maximum whose value is larger than all subsequent local maxima? Not possible unbounded. However, we can choose a local maximum that is larger than the next two local minima, etc. Maybe use a Ramsey/Erdos-Szekeres on the sequence of local extrema to find a monotone AP of length 3 with larger d. For d=1, signs alternate. Let even positions be local maxima (WLOG). Then the subsequence of even positions is a zigzag; its local maxima/minima correspond to d=2 extrema. This hierarchy is like a binary tree. The values at nodes of the tree (positions congruent modulo 2^r) have extremum properties. Since values are integers, maybe by König's infinity there is a node whose value is between its ancestor and descendant, yielding monotone AP. This could be a proof of no 3-free permutation of Z. But finite 3-free permutations of arbitrary length contradict a purely finite Ramsey; the obstruction must use infinite in both directions and unboundedness of Z. Consider the tree of dyadic APs. For each finite binary string (residue class), the values at positions in that class form a zigzag sequence (for step 2^r). The extremum pattern at level r+1 refines. This is analogous to a binary tree of sequences. A 3-free permutation would assign distinct integers to nodes (integer positions) such that each sequence is zigzag. This is possible for finite depth and finite number of nodes. For infinite depth and infinite nodes in both directions, can we assign integers? This is like a 2-adic tree labeling problem. Maybe impossible by a Baire category/compactness? The tree has countably many nodes (all integers). We need label each integer with a distinct integer such that for each node (position) and each parent? The condition for all d is equivalent to: for each d, the sequence along cosets mod 2d is zigzag. This is a set of local constraints. It may be possible to label by a 2-adic order of the tree. In fact, the 2-adic dense order labels the 2-adic tree densely. To label with integers (discrete) might be impossible because the tree is infinite in both directions and each branch corresponds to an AP with infinite nodes; a zigzag labeling of an infinite branch with integers unbounded both may be possible (e.g., ... low, high, low, high). But consistency across branches may force a dense order. Let's attempt a direct construction of a 3-free labeling of Z using the 2-adic tree. We can label each integer n by a real number equal to its 2-adic reversed binary sequence (a point in [0,1]). This gives a 3-free order (dense). Then we need to map these real numbers to integers (a bijection) while preserving the property that for each AP, the middle label is extreme relative endpoints. If we choose a monotone bijection from the dense order to Z? No monotone bijection from dense countable order to Z exists (order types differ). But we don't need to preserve all order relations, only betweenness for AP triples. The dense order's AP betweenness is a partial order; maybe it has a linear extension of type Z. The dense order is a linear extension; we need another. The partial order might be "interval" type that admits discrete extensions. This is plausible. Let's test finite: The 3-free condition on a finite set does not determine a unique order; there are many linear extensions. For the full 2-adic partial order on Z, do there exist linear extensions of type Z? This is a question of whether the poset has a discrete linear extension. Many countable posets have dense and discrete linear extensions. For example, the poset with no comparabilities has type Z extension. Our poset is highly structured but maybe thin. Can we use a theorem: Every countable poset of width? has a linear extension of type Z? Not true (e.g., Q with usual order has only dense linear extensions, no Z extension). Our poset might be Q-like. The 2-adic order partial order may be dense enough to be Q-like, forbidding discrete extensions. We need know. Maybe the 3-free conditions imply that between any two elements in the order there is another element (density), even in any linear extension. If so, no permutation. Let's try to prove density: In any 3-free order (not necessarily type Z) on Z, for any a b. Let's derive more using the APs involving a+2d and a-d. Consider AP (a+2d, a, a-2d): middle a. We know a+2d a > a-2d? Actually a-2d b. For a-d max, need a-3d < a-d. (no strong) Consider AP (a-2d, a, a+2d) done. Consider AP (a+d=b, a+2d, a+3d): middle a+2d. We know a+2d < a < b. Endpoint b > a+2d. For a+2d extreme, since b>a+2d, to avoid b> a+2d > a+3d? Wait middle a+2d. Endpoints b (above) and a+3d unknown. For a+2d to be extreme, it must be below both endpoints (min), so a+3d > a+2d. (4) Thus a+3d > a+2d. Could be above b or below a. Consider AP (a+2d, a+3d, a+4d): middle a+3d. If a+3d > a+2d. To avoid increasing if a+4d > a+3d, need a+3d > a+4d (max). So a+4d < a+3d. (5) Now AP (a, a+2d, a+4d): middle a+2d. We know a+2d < a. Endpoint a. To avoid decreasing a > a+2d > a+4d, need a+4d > a+2d (min). Combined with (5): a+2d < a+4d < a+3d. (6) AP (a+2d, a+4d, a+6d): middle a+4d. Since a+2d a+6d < ... all maybe below a? AP (a, a+4d, a+8d) with a+4d? If a+4da+2d, no relative a. If a+4d > a, then since b=a+d is immediate successor of a, a+4d cannot be between a and b; it could be >b or b, then even subsequence crosses above b. Possible. What about negative side: AP (a-d, a-2d, a-3d): middle a-2d < a. Endpoint a-d > a. For a-2d extreme, since a-d > a > a-2d, it must be min, so a-3d > a-2d. (7) AP (a-2d, a-3d, a-4d): a-3d > a-2d; to avoid increasing need a-4d < a-3d. (8) AP (a-4d, a-2d, a): middle a-2d < a. Endpoint a > a-2d. To avoid decreasing a > a-2d > a-4d, need a-4d > a-2d. Combined: a-2d < a-4d < a-3d. Symmetric. Now, consider the point a+3d. From (4), a+3d > a+2d. Could a+3d be between a and b? No adjacent. If a+3d > b, OK. If a+3d < a, OK. The AP (b, a+2d, a+d?) no. Use the fact that d=b-a. The point b-d = a? trivial. The point b+d = a+2d < a. The point b-d? a. The reflection of b around a is a-d > b. The reflection of a+2d around a is a-2d < a. The order has pairs (a+2d,a-2d) below a, (a-d,a? ) above. Maybe consider the midpoint of b and a+2d: (a+d + a+2d)/2 = a + 1.5d, not integer if d odd. If d even, midpoint integer m=a+3d/2. Could use d's parity. If d is even, let d=2e. Then b=a+2e. The AP (a, a+e, a+2e=b) has middle a+e. Since ab. Thus for every even d, the midpoint a+d/2 is outside the adjacent order interval. If d has a large power of 2, there are many midpoints. Maybe one of them is forced into the interval by other constraints. If d is odd, no midpoint. But we can consider d+? Maybe there is always a multiple? We can choose an adjacent pair with d even? Is there guaranteed? The successor map S(a)=a+d_a. The differences d_a are integers. Could all d_a be odd? If S(a) parity differs from a for all a, then S maps even to odd and odd to even. A bijection of Z with one Z-chain and parity-flipping is possible (e.g., S(n)=n+1). Identity order parity flips but not 3-free. So not guaranteed. If d even, let e=d/2. The midpoint m=a+e. Since a,b adjacent, m is outside (a,b) in order. Suppose mb, then a max? OK. So midpoint outside is consistent. If d has 2^r factor, consider the binary subdivision points a + k d/2^r. The order must place them outside the interval (a,b) in a 3-free pattern. Maybe as r→∞, infinitely many domain points (a + d * k/2^r, only integer if 2^r|d finite) no, d fixed finite, only finitely many integer midpoints. Not infinite. But we can apply to the point a+2d (below a). Its successor in order might have some difference, leading to an infinite descent into the adjacent interval? Maybe. Let's search for a known proof of "no 3-free permutation of Z". If it exists, it might be short: Choose x such that x+1 is the immediate successor of x in the permutation (i.e., π^{-1}(x+1)=π^{-1}(x)+1?). Wait "immediate successor in the permutation order" means values x and x+1? Maybe by a pigeonhole/compactness, there must be two consecutive integers in value that are adjacent in the permutation order? In a permutation of Z (values all integers), consider the pairs (n,n+1) in value order. In the position order, are there adjacent pairs? We need two values that are consecutive integers and consecutive positions? Not guaranteed. But maybe by a theorem on permutations of Z, there exist x,y such that |x-y|=1 and positions adjacent? No, e.g., π(n)=2n? not bijective. A permutation can avoid consecutive values adjacent: π(n)=n+(-1)^n floor? Maybe possible. If there is an adjacent pair in the permutation order with value difference 1, then the earlier d=1 argument may yield contradiction? Let's test. Suppose values a and a+1 are adjacent in the permutation order (positions consecutive). If a+1 is immediate successor of a in value order? Wait we need adjacent in the order by values, not domain. In a 3-free permutation (APs on positions, values monotone), the order by values is the usual integer order; adjacent values are consecutive integers. If two consecutive values are adjacent in the permutation order? The 3-free condition is on APs in positions, not values? I'm mixing. Let's fix the interpretation that APs are on positions and values monotone. Then a permutation π:Z->Z. A monotone 3-AP is positions n,n+d,n+2d with values monotone. The order by values is the usual order on π(n). The 3-free condition: for all n,d, π(n+d) extreme in value order relative π(n),π(n+2d). This is a sequence π indexed by Z with no monotone 3-AP. The "order type Z" is on the index side (positions), not on the value side? The values are Z usual order, discrete. The sequence positions are Z usual order, discrete. The 3-free condition couples two discrete orders. In our order-theoretic reformulation, we considered a total order < on the domain Z by value; its order type is Z (values). The APs are on the domain usual order. That's correct. Adjacent in the value order are consecutive values, but their domain points are some pair a,b with b? The immediate successor in the value order of a is a+1 (consecutive integer), not an arbitrary domain point. Ah! I was treating the order < on the domain by value, where immediate successor of a (domain point) is the domain point whose value is one more than a's value. Since values are all integers, if a has value r, its immediate successor in the order is the domain point with value r+1. The difference d = s(a)-a in domain can be arbitrary. Good. Adjacent pair in value order always has value difference 1, domain difference d. We used that. Now, in the value order, the immediate successor of the domain point a is the domain point s(a) with value value(a)+1. The AP (a, s(a), 2s(a)-a) is an AP in the domain (positions) if s(a) and a are integers (yes). The 3-free condition applies. Our derived constraints hold for any adjacent values (consecutive values) in a 3-free permutation. Thus if a 3-free permutation of Z (APs on positions) exists, for every value r, if the positions of values r and r+1 are at domain points a and b, then b-a=d and the constraints hold. Can we use the fact that value differences are 1 to get a contradiction via iterating values? The successor map in the order is r -> r+1 (value), but in domain it maps a to b. The constraints involve domain points a+2d, a-d, etc., which correspond to other values (not necessarily consecutive). Maybe we can follow the value chain: value r at a, r+1 at b=a+d, r+2 at some c. The AP (a,b,c?) Not necessarily c=a+2d; c is position of value r+2, unknown. The 3-free condition for the domain AP a,b,a+2d involves value at a+2d, which is some value not r+2 generally. Hard. But maybe we can choose r such that the positions of three consecutive values form an AP? If values r,r+1,r+2 are at positions forming an AP, then they would be monotone (consecutive values) and forbidden. Thus for every r, the positions of r,r+1,r+2 are not in AP with middle r+1. This is a different condition. Could use this to prove existence? Not. Let's switch interpretation to APs on values, positions monotone. Then adjacent values in domain are consecutive values, and APs are on domain values; the 3-free condition is similar with domain and range swapped. The adjacent pair in the permutation order (positions) may not be consecutive values. Ambiguity again. Maybe we should look for the original problem context. "permutation of ℤ" and "monotone k-term arithmetic progression x1 <⋯ < xk" The notation x_i likely are the terms of the permutation, i.e., values, not positions. A "monotone arithmetic progression" x1<...Z 3-free question. Good. Now, can we find known result for this exact phrasing? "monotone k-term arithmetic progression" in a permutation often means the indices (positions) are monotone? Let's search memory: In the context of "permutations and arithmetic progressions", a "monotone arithmetic progression" in a permutation π of [n] is a sequence π(i), π(i+d), ..., π(i+(k-1)d) that is monotone. They denote the terms x_i = π(i+...)? Hmm. But the problem writes x_1<... values). For AP constraint on values, we need the inverse. But if the sequence is symmetric, maybe. Let's test the sequence candidate for no monotone 3-AP in positions (APs on positions, values monotone). Sequence a_n (n∈Z) maybe: ... , -8, 3, -14, 4, -20, 5, -6, 2, -12, -18, 1, -24, 6, ... Not sure. Maybe the 3-free permutation of Z is given by the following rule (from memory of a MathOverflow answer by Noam Elkies?): a(n) = (-1)^n 2^{v_2(n)} (n+1) for n≠0, and a(0)=0? Let's test with v_2(0)=∞. For n=1: -2, n=2: +? v2(2)=1, 2*(3)=6, n=3:-4, n=4: v2=2,4*5=20, n=5:-6, n=6:2*7=14, n=7:-8, n=8:8*9=72. Sequence -2,6,-4,20,-6,14,-8,72. d=1 zigzag? -2<6>-4<20>-6<14>-8<72 yes. d=2: even positions -2,-4,-6,-8 monotone decreasing (bad). If use v_2(n+1): n=1: -? v2(2)=1,2*? maybe -? We tried. What formula makes even subsequence zigzag? Need the sign/magnitude for n=2m to be a 3-free sequence in m. If a(2m) = -2 a(m) (negative scaling), then even subsequence is a scaled copy (with sign) of a(m), preserving 3-free. Similarly odd subsequence. For d=1, we need a(2m+1) extreme between a(2m) and a(2m+2). If a(2m)=α a(m), a(2m+2)=α a(m+1), and a(2m+1)=β a(m)+c. To be extreme for all m, choose β large sign relative. Suppose a(2m)=2a(m), a(2m+1)=-2a(m)-1. If a(m) and a(m+1) have the same sign, odd extreme. If not, fail. But if the base sequence a(m) is such that adjacent terms always have the same sign? Impossible for a permutation of Z, but perhaps we can arrange that a(m) is positive for m in certain blocks and negative for other blocks, and the recurrence's sign handles? For d=1, the condition only involves a(m),a(m+1). If we ensure a(m) and a(m+1) have the same sign for all m, then a(m) sign constant, impossible. But if a(2m) and a(2m+2) are both scaled by 2, their signs are signs of a(m),a(m+1). If they differ, the odd value -2a(m)-1 might still be extreme depending on magnitudes. Let's analyze: Let A=a(m), B=a(m+1). Even endpoints 2A,2B. Odd x=-2A-1. If A and B have opposite signs, say A>0, B<0. Then endpoints positive 2A and negative 2B. x is negative (if A>0). Is x extreme? It could be less than both if x < 2B (more negative than 2B), or greater than both if x>2A (impossible). So need -2A-1 < 2B, i.e., -A-0.5 < B. Since B negative, this means |B| > A+0.5. Thus if the negative B has larger magnitude, odd min extreme. If B's magnitude not large enough, fail. If A<0, B>0, x positive; need x > 2B? -2A-1 > 2B => |A| > B+0.5. So odd extreme if the sign of a(m) has larger magnitude than a(m+1) when signs differ. This is a condition on adjacent magnitudes. We can design a(m) to satisfy recursively: when adjacent signs differ, the one with the sign of the odd child has larger magnitude. Maybe a simple sequence where magnitudes strictly decrease as m increases? But domain Z unbounded both, cannot. For one-sided m≥0, magnitudes could decrease to 0, impossible distinct positive. For Z, could alternate. Consider the recurrence: a(2m) = 2 a(m) a(2m+1) = -2 a(m) - 1 This maps Z to all integers if a is a bijection? Ranges even (2Z) and odd negative? Wait -2a(m)-1 is odd (all odd as a(m) ranges Z). Union Z. Invertible: given y, if even y=2x; if odd y=-2x-1 => x=-(y+1)/2. So this is a bijection on Z independent of a? Actually if a is a permutation, the recurrence defines a new sequence a' of length? For infinite Z, the map T on functions: (Ta)(2m)=2a(m), (Ta)(2m+1)=-2a(m)-1. If a is a bijection, is Ta a bijection? The ranges of 2a(m) and -2a(m)-1 are all evens and all odds, disjoint, union Z. Yes, Ta is a bijection for any bijection a. We can iterate: start with a_0(n)=n (identity), define a_{r+1}=T a_r. Maybe the limit? Let's compute a_1: n=0: a(0)=0 1: -1 2: 2a(1)=-2 3: -2a(1)-1=1 4: 2a(2)=4 5: -2a(2)-1=-3 6: 2a(3)=2 7: -2a(3)-1=-1 duplicate? a1(7)=-1 duplicate. Wait if a_0 is identity on Z, the recurrence may not be injective? We argued ranges partition but values depend on a(m); for each y even, preimage m with a(m)=y/2 unique; for odd, preimage m with a(m)=-(y+1)/2 unique. Since a is bijection, each y has unique m for its parity, so Ta is bijection. But I computed duplicates because I didn't consider negative indices? For n=1, a0(1)=1 -> a1(1)=-3? Wait formula a(2m+1)=-2a(m)-1. For n=1, m=0: -1. n=3, m=1: -2*1-1=-3, not 1. I mistakenly used a(m)=? Let's recompute a1 for n≥0 with a0(n)=n: a1(0)=0 a1(1)=-1 a1(2)=2*a0(1)=2 a1(3)=-2*a0(1)-1=-3 a1(4)=2*a0(2)=4 a1(5)=-2*a0(2)-1=-5 a1(6)=2*a0(3)=6 a1(7)=-2*a0(3)-1=-7 So a1: 0,-1,2,-3,4,-5,6,-7,... This is zigzag? 0,-1,2,-3,4,-5: d=1: 0>-1<2>-3<4>-5 good. d=2: even subsequence 0,2,4,6 monotone increasing (bad). But a1 is just parity sign with increasing magnitude. Now apply T again: a2(2m)=2a1(m), a2(2m+1)=-2a1(m)-1. Compute a2: 0:0 1: -2*a1(0)-1=-1 2: 2*a1(1)= -2 3: -2*a1(1)-1 = 3 4: 2*a1(2)=4 5: -2*a1(2)-1 = -5 6: 2*a1(3)= -6 7: -2*a1(3)-1 = 5 8: 2*a1(4)=8 9: -2*a1(4)-1=-9 10:2*a1(5)=-10 11:-2*a1(5)-1=9 Sequence: 0,-1,-2,3,4,-5,-6,5,8,-9,-10,9,12,-13,-14,13,... Check d=1: 0,-1,-2 decreasing (bad). So iterating T worsens. What if T alternates sign differently: a(2m)=2a(m), a(2m+1)=2a(m)+1? identity. a(2m)=2a(m), a(2m+1)= -2a(m)+1? We tried. Maybe T with a(2m)=2a(m)-1, a(2m+1)=2a(m)+1? Let's test injectivity: ranges odd? 2a-1 and 2a+1 are odd, not partition evens. No. Need partition into two sets that are separated in a way. The finite bit-reversal recursion can be written as for a finite interval of values [L, L+2^m): even indices get lower half, odd indices get upper half. For infinite, we can use a family of value intervals I_m that are not globally separated but locally. The operator T above used ranges evens/odds, not separated. Maybe use a(2m) = 2 a(m), a(2m+1) = 2 a(m) + 2^{v_2(m)+1}? The offset should be larger than the range of differences in a over neighboring m. If a(m) is a 3-free permutation with bounded local differences? Not. Let's consider a simpler goal: construct a permutation with no monotone 4-AP, assuming 3 unavoidable. This may be easier and known. The largest k would be 3. We can perhaps provide a construction using block sizes growing rapidly. Let's explore that. We need a permutation of Z (bijection) such that every monotone AP has length ≤3 (i.e., no length 4). If 3 is unavoidable, answer 3. Construction: For any finite k, can avoid k? Maybe for Z, can avoid arbitrarily long monotone APs? If so largest k might be 2? Wait if 3 unavoidable but can avoid 4, answer 3. Can we avoid arbitrarily long? Maybe by the 2-adic dense order approximations, one can avoid 4 but not 3. There may be a general theorem: For any k≥4, there is a permutation of Z with no monotone k-AP, but every permutation has monotone 3. This would make largest 3. This is plausible and elegant: 3 is the threshold. Can we construct a permutation of Z with no monotone 4-AP? Perhaps using the 2-adic order dense (which has no 3) and then "discretize" it to a permutation; the discretization may introduce 3 but cannot create 4 if done carefully. Or use a sequence of blocks with bit-reversal of length 3-free, and separate blocks so any 4-AP crossing blocks is nonmonotone. This is a standard construction for avoiding long monotone APs in permutations of Z. Let's attempt a block construction for no 4-AP (APs on values, positions monotone, or positions AP? Need choose). Suppose we build a permutation q:Z->Z (value -> position) with no monotone 4-AP in values. Partition values into blocks B_m of consecutive integers (in value order) of size N_m, and assign their positions in intervals P_m of consecutive positions, with the property that any 4-term AP of values that crosses blocks has its positions not monotone. If the blocks are arranged with alternating order (some blocks increasing, some decreasing) and sizes growing fast, maybe. For APs on values, a 4-term AP has values a,a+d,a+2d,a+3d. If d is large, it crosses many blocks. If block size N_m << d, then each block contains at most one term of the AP. The positions of these terms are in different position blocks P_m. If the position blocks P_m are arranged in an order that is not monotone along the value AP, we can avoid. For example, make P_m alternate high/low with m in a 2-adic pattern. If d is small, AP stays within or near a few blocks; we can handle by making each block's internal permutation 3-free (no monotone 3) so certainly no 4 internally? Wait if internal no 3, then no 4 inside. Cross-block with small d might involve at most a few blocks; need handle. This is getting complex but possible. The problem likely expects a known result, not a full construction in answer? We need provide final answer with enough reasoning. Let's see if there is a simpler construction avoiding 4 using the sequence q(n) = floor(n/2) with alternating blocks? For APs on positions (not values), maybe easier: Sequence a_n with no monotone 4-AP can be built by blocks of length 3? But 3 unavoidable? If no 3-free, maybe a simple sequence with 3 but no 4: e.g., repeat finite 3-free blocks with large gaps. Let's test for APs on positions: Let a_n be a sequence of blocks, each block is a 3-free permutation of an interval, and blocks are separated in value by large gaps and ordered alternately. Any 4-term AP in positions either lies within one block (impossible if block 3-free) or crosses blocks. If it crosses, at least two terms in different blocks. With large gaps, the values in intermediate blocks may be extreme? Need ensure nonmonotone. For length 4, if an AP has positions crossing blocks, the middle two positions might be in blocks with values far from endpoints, possibly monotone? Could arrange block values alternately high/low so any cross-block 4-AP has a local extremum at the middle. This is plausible. If the answer is 3, we also need prove 3 unavoidable. Is there a simple proof? Let's search memory: "Every permutation of the integers contains a 3-term monotone arithmetic progression" might be proved by considering the first three terms? Wait for a permutation of Z indexed by Z, maybe use the fact that the sequence is two-sided and values unbounded both. Suppose no monotone 3. Then signs alternate for d=1. Consider the sequence of local maxima (say odd positions). It is a bi-infinite sequence of integers. By the Erdős–Szekeres theorem for sequences of integers? Any bi-infinite sequence of distinct integers has an infinite monotone subsequence? Yes, but we need AP indices. The local maxima are at odd positions; their subsequence (over m) has no monotone triple? For d=2, the local maxima/minima of odd subsequence alternate. This hierarchical zigzag may be impossible over Z because of a compactness/well-ordering argument? But finite examples exist. Maybe the proof uses the fact that the set of local maxima has no maximum/minimum, and one can find a monotone 3-AP with larger d by a diagonal argument. This is nontrivial. Let's try to prove 3 unavoidable for q:Z->Z (APs on positions). Suppose no monotone 3. For each d, the d-decimated sequence is zigzag. In particular, for each d, the sequence q(n+kd) has signs σ(n,d) alternating. This implies for each d, the set of local maxima/minima in that decimation is infinite both directions. Can we use the infinite pigeonhole to find d and n such that q(n),q(n+d),q(n+2d) monotone? That's what we assume false. So sign pattern exists. Maybe sign pattern can be realized by a total order? We were unsure. Let's search online memory: "3-free permutation of the integers" I think I saw a concrete sequence in a paper "On 3-free permutations" by R. P. Stanley? It says: There are no 3-free permutations of the positive integers. However, there is a 3-free permutation of the integers, namely ..., -5, 2, -8, 3, -11, 4, -2, 0, -6, 5, -10, ... Let's test this sequence for APs on positions (the sequence itself). Sequence positions maybe ... let's index by n∈Z: n: ... -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10? Values: ... -8? Let's use the snippet: ... , -5, 2, -8, 3, -11, 4, -2, 0, -6, 5, -10, ... Suppose this is the sequence a_n with n increasing left to right. Let's assign n=0 value 0 maybe: ... -2,0,-6,5,-10,...? The snippet includes 0. Need full. It seems to have pairs (positive, negative) with magnitudes growing. Check a small window: -2,0,-6,5,-10. d=1: -2,0,-6 non (0 max); 0,-6,5 non (-6 min); -6,5,-10 non (5 max). d=2: -2,-6,-10? middle -6? -2>-6<-10? -6 is max? -6 > -10 but < -2? Actually -6 < -2, so -2 > -6 > -10 decreasing (monotone) bad if these are consecutive even indices. So not 3-free for positions. Maybe APs on values? The sequence lists values in position order; APs on values would look at positions of values. Hard. Maybe the sequence is the order of values by positions (a permutation of Z) and is 3-free in the sense APs on positions? The snippet fails, so maybe I misremember. Let's use a computational heuristic: Can we find a periodic or simple 3-free sequence of length, say, 16 with values Z (not necessarily permutation) by solving constraints? We have finite bit-reversal length16: 0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15. This is 3-free on positions for that finite set. If we want an infinite sequence, we could use a "limit" where values are not 0..2^m-1 but shift each prefix. Perhaps we can take the prefix of length 2^m as bit-reversal of a transformed set S_m such that S_{m+1} contains S_m. Is there a nested family of 3-free permutations? Let's try to find one by choosing values from Z recursively. We want a sequence a_n (n≥0) such that for each m, the first 2^m terms are 3-free. We can choose a_{0..15} as bit-reversal scaled? Need extend to length 32 while preserving first16. Is it possible? Let's test with a general method: Given a 3-free sequence S of length N, can we extend to length 2N by appending N new terms with values chosen from a new interval far above/below, using a transformation? For finite bit-reversal, the length 2N sequence is not an extension but a rearrangement. Maybe we can choose S_{2m} and S_{2m+1} as interleavings of two copies of S with a large offset, while keeping old prefix? If we keep old prefix, the old indices' values cannot be changed. Our greedy attempts suggest some finite 3-free prefixes are not extendable indefinitely. But maybe there exists a nested sequence. Let's try to construct length 32 extending length16 bit-reversal? The first16 values 0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15. At n=16, constraints from d=1..8. The allowed interval for a16 might be large if we choose future? It depends only on previous values. Let's compute quickly for n=16 (0-based). Need for each d=1..8, L=a_{16-2d}, M=a_{16-d}, require a16 on same side of M as L. d=1: L=a14=7, M=a15=15, L x<15. d=2: L=a12=3, M=a14=7, L x<7. d=3: L=a10=5, M=a13=11, L x<11. d=4: L=a8=1, M=a12=3, L x<3. d=5: L=a6=6, M=a11=13, L x<13. d=6: L=a4=2, M=a10=5, L x<5. d=7: L=a2=4, M=a9=9, L x<9. d=8: L=a0=0, M=a8=1, L x<1. Intersection x<1. Unused values <1: negative integers. Choose a16=-100. n=17: d=1: L=15,M=-100 => L>M => x>-100. d=2: L=7,M=15? indices13? n=17,d=2: L=a13=11? Wait n-2d=13, M=a15=15. L=11 x<15. d=3: L=a11=13, M=a14=7 => L>M => x>7. d=4: L=a9=9, M=a13=11 => x<11. d=5: L=a7=14, M=a12=3 => x>3. d=6: L=a5=10, M=a11=13 => x<13. d=7: L=a3=12, M=a10=5 => x>5. d=8: L=a1=8, M=a9=9 => x<9. Intersection: max( -100,7,3,5 ) =7, min(15,11,13,9)=9 => 7 -1; d=3 x>7; d=5 x>3; d=7 x>5 => x>7. Upper bounds: d=2 x<15; d=4 x<11; d=6 x<13; d=8 x<9 => x<9. So x must be 8 (used) regardless of a16 as long as a16<-1? If a16=-1, d1 lower x>-1, still lower 7. If a16=0? used. So impossible. Thus the bit-reversal prefix of length16 is dead. Maybe no finite prefix of a 3-free sequence can be extended to arbitrary length? But there are arbitrarily long finite 3-free sequences (bit-reversal of length 2^m). They are not nested; each length has a different prefix. To get infinite, we need a nested chain of finite 3-free sequences where each prefix extends the previous. The tree of finite 3-free sequences may have arbitrarily long branches but no infinite branch? By compactness of infinite branching? It could have no infinite branch if the tree is not finitely branching and not closed under pruning? Actually if there are arbitrarily long branches in a rooted tree, there is an infinite branch if every node on a long branch has at least one extension to arbitrary length. The bit-reversal length16 has length16 branch but cannot extend to17; there may be another length16 prefix that can extend to17 and beyond. We need find a pruned tree. Our earlier greedy with values like 0,1,-2,-1,100,2,... got to length7 and seemed extendable if choose wisely. Maybe there is an infinite branch. Let's try to continue that sequence with a strategy. Sequence so far (n=0..6): 0,1,-2,-1,100,2,? We chose n=5=2, n=6? Let's restart with a more systematic strategy to keep allowed intervals wide. We need a_n distinct integers. Let's use a heuristic: At step n, if allowed interval is (L,U), choose a_n near the middle? Or choose extreme to keep future intervals wide? Our sequence: a0=0 a1=1 (choose >0) a2 must <1; choose -2 (far below) a3 must >-1? from d1: L=1,M=-2 => x>-2. Choose -1 (just above -2) to keep d2 same direction? We chose -1. a4 allowed >-1; choose 100 (far above) a5 allowed -1 L>M => x>2. d2: L=a2=-2,M=a4=100 => L x<100. d3: L=a0=0,M=a3=-1 => L>M => x>-1. Allowed 2 if 2 new a7 < x. d2: L=a3=-1,M=a5=2 => L a7<2. d3: L=a1=1,M=a4=100 => L a7<100. d4: L=a? n-8=-1 none? d≤3 for n=7? 2d≤7 => d=1,2,3. So upper bounds: a7<2 from d2 (strong), a7 L>M => a8 > -1000. d2: L=-1,M=x => L-1) => a8 < x. d3: L=0,M=-1? indices2? n=8,d3: L=a2=-2? wait n-2d=2, M=a5=2. L=-2 a8<2. d4: L=a0=0,M=a4=100 => L a8<100. Allowed -1000 < a8 < min(x,2,100)=2. Available integers <2: many (3? no, 2 used, 0,1 used, negatives). Choose a8=3? 3<2? no. Choose a8=-1001? allowed >-1000? no if -1001. Choose a8=-999 (unused) maybe. But need keep future. Let's choose a8= -999. Sequence: 0,1,-2,-1,100,2,x,-1000,-999,... with x between2,100. Let's pick x=50. Check n=9 constraints: d1: L=a7=-1000,M=a8=-999 => L a9 < -999. d2: L=a5=2,M=a7=-1000 => L>M => a9 > -1000. d3: L=a3=-1,M=a6=50 => L a9 <50. d4: L=a1=1,M=a5=2 => L a9<2. Allowed -1000 < a9 < -999: no integer! Fail. If choose a8 differently? At n=8 allowed a8<2 and >-1000. The d1 at n=9: L=a7=-1000, M=a8. If a8 > -1000, L a9 < a8. d2: L=a5=2, M=a7=-1000 => a9 > -1000. d4: a9<2. Thus allowed (-1000, min(a8,2)). To have an integer, need a8 ≤ -999? If a8=-999, interval (-1000,-999) empty. If a8=-1000? not allowed (distinct and >? a8 > -1000? from n8 d1 yes a8>-1000). If a8=-1000? no. If a8=-1001? not allowed. To have an integer, need a8 ≤ -1000? impossible. Wait if a8 is much smaller, say -2000, but n8 required a8>-1000 due d1 (L=100? Let's recompute n8 d1: L=a6=50, M=a7=-1000. L>M => a8 > -1000. Yes lower bound -1000. Thus any a8 > -1000, and n9 d1 gives a9 < a8, d2 gives a9 > -1000, so a9 must be between -1000 and a8, an integer if a8≥-999. If a8=-999, none; if a8=-998, a9=-999 available! I chose -999, should choose -998. Let's set a8=-998. Then n9 allowed -1000 a9=-999. Good. Sequence n=9=-999. n=10 constraints: d1: L=a8=-998,M=a9=-999 => L>M => a10 > -999. d2: L=a6=50,M=a8=-998 => L>M => a10 > -998. d3: L=a4=100,M=a7=-1000 => L>M => a10 > -1000. d4: L=a2=-2,M=a6=50 => L a10 <50. d5: L=a0=0,M=a5=2 => L a10<2. Allowed max(-999,-998,-1000)=-998 < a10 <2. Available integers: -997,...,-1000? -1000 used, -998 used, -999 used; choose -997. a10=-997. n=11: d1: L=-999,M=-997 => L a11 < -997. d2: L=50,M=-999 => L>M => a11 > -999. d3: L=100,M=50? indices5? n=11,d3: L=a5=2? wait n-2d=5, M=a8=-998. L=2 > M=-998 => a11 > -998. d4: L=-1,M=2? indices3? n-8=3, M=a7=-1000. L=-1 > -1000 => a11 > -1000. d5: L=1,M=50? indices1,6: 1<50 => a11<50. Allowed > -998 and < -997: no integer. Fail. If choose a10 differently? At n10 allowed -998 < a10 <2. To avoid n11 empty, need? n11 d1: L=a9=-999, M=a10. If a10>-999, L a11 < a10. d2: L=a6=50, M=a8=-998 => a11 > -998. Thus need -998 < a11 < a10, so a10 ≥ -997. If a10=-997, empty. If a10=-996, a11=-997 available. So choose a10=-996. Then a11=-997. Pattern: we are creating descending blocks of negative integers by pairs. This might continue! Let's set: a0=0,a1=1,a2=-2,a3=-1,a4=100,a5=2,a6=50,a7=-1000,a8=-998,a9=-999,a10=-996,a11=-997,... It seems we can use large positive values to create wide intervals, then fill negative integers in a pattern. Maybe an infinite construction exists. This suggests 3-free sequence of Z may exist! The greedy with careful choices can avoid dead ends by choosing values at least two steps away to leave room. This is reminiscent of a general strategy: when constraints force an interval (A,B), choose the endpoint leaving at least two unused integers for the next step. Since values are unbounded, we can always do this. We also need surjectivity; we can interleave all integers by priority, using large positive/negative reservoirs. This suggests existence of a one-sided 3-free sequence of Z is plausible. If so, a permutation of Z indexed by N exists with no 3. For a permutation of Z indexed by Z, we can take a one-sided 3-free sequence of Z and extend to both sides? Maybe easier: we can build a bi-infinite sequence by doing the same greedy on both sides, or take a one-sided sequence and place it in the nonnegative positions, and another in negative positions with a separation? But APs crossing 0 need handling. However, if we have a one-sided 3-free sequence of Z, we can make a two-sided permutation by using two copies (scaled) on even/odd? Need ensure cross APs. But maybe the problem's permutation of Z is one-sided (sequence indexed by N). The notation x_1<...x, require new y > x. d=2: L=a13=11, M=a15=15 -> y<15. d=3: L=a11=13, M=a14=7 -> y>7. d=4: L=a9=9, M=a13=11 -> y<11. d=5: L=a7=14, M=a12=3 -> y>3. d=6: L=a5=10, M=a11=13 -> y<13. d=7: L=a3=12, M=a10=5 -> y>5. d=8: L=a1=8, M=a9=9 -> y<9. Intersection: y>max(x,7,3,5)=7 (since x<1), y<9. Integer y=8. But 8 is already used (a2=8). If we had not used 8, could choose 8. The prefix used all 0..15, so 8 used. If at n=16 we choose x=8? not allowed x<1. Thus no extension. So not every finite 3-free sequence is extendable. But maybe every finite 3-free sequence that is "cofinal" (uses a consecutive interval of values?) is extendable? Bit-reversal uses 0..15; not. If we use values with gaps, maybe extendable. The tree of extendable sequences may be nonempty. Can we prove there is an infinite extendable sequence by a compactness argument with the property "extendable to any length"? The set of finite sequences that are extendable to arbitrary length is nonempty for each length (bit-reversal length N is extendable to length N but not beyond; need extendable to arbitrary length). We need a sequence that is 3-free of arbitrary finite length, not necessarily nested. By König's lemma for countable infinite branching, if for each N there exists a 3-free sequence of length N, does there exist an infinite 3-free sequence? Not in general, but for local constraints on a discrete alphabet with the ability to choose new values far away, perhaps yes. We can use a diagonal construction: choose a_0 arbitrarily. Suppose we have chosen a_0,...,a_{n-1} with the property that for every m≥n there is a 3-free extension of length m. We need choose a_n preserving this property. The set of possible a_n that allow extensions to length m may be finite? We need show intersection over m of infinite sets nonempty. This is a compactness in Z with order topology? Z not compact, but we can use the fact that allowable values lie in an interval with infinitely many integers; maybe. Alternatively, use a known theorem: The set of 3-free permutations of Z (as a closed set in the product topology of S_Z?) is nonempty because the constraints are compact? The space of permutations of Z (bijections) is a Polish space; the property "for all n,d, a_{n+d} is not between" is G_delta? Nonempty if finite satisfiable and the alphabet is compact. Z is not compact, but we can one-point compactify. Finite satisfiability with unbounded values may imply infinite by the compactness theorem for logic with infinite language? We can treat each integer value as a constant and use first-order logic? We want a sequence a_n with distinct integer values (not predetermined). We can use the compactness theorem: For each finite set of constraints (finiteness of indices and values), there is a finite model? Not straightforward. Maybe there is a known result: There exists a 3-free permutation of the integers, and the proof uses the compactness theorem. If so, answer 2. The question might be from someone aware of this and asks largest k; answer 2 with reference to 3-free permutations. But they might expect a construction, not just compactness. Let's search memory of "3-free permutation of the integers" exact: I think the answer is: There is no 3-free permutation of the positive integers, but there is a 3-free permutation of the integers. In fact, the 2-adic ordering gives a 3-free permutation of the integers. Wait "permutation" might mean a total ordering, not a sequence. The 2-adic ordering of the integers (which is dense) is not a permutation in the usual sense, but in combinatorics on words, a permutation of a set is just a bijection from N? Hmm. Let's consider the possibility that the intended answer is "2" and the construction is simply: Use the 2-adic order. They might not require the permutation to be indexed by Z in the usual order; a permutation of Z can be viewed as a bijection from Z to Z, and the 2-adic order gives a bijection? The 2-adic order is a linear order on Z, but not of type Z. However, a linear order on a countable set is not a permutation unless it is order-isomorphic to Z? If they loosely call any bijection a permutation, the 2-adic order is not a bijection but an order. To get a bijection, you can assign ranks in the 2-adic order; but the ranks would be a dense order, not integers. So no. Maybe there is a discrete 2-adic order on Z after all. Let's re-examine the 2-adic order on Z defined by reversed 2-adic bits. We thought it's dense because powers of 2 approach 0. But if we include negative integers (eventually 1 sequences), maybe the order is not dense? The set of eventually 0/1 binary sequences: Is it dense? Take two sequences ss, eventually 0. Yes. If s tail not all 0/1? All sequences in set are eventually 0 or 1, but can have arbitrary finite prefix. Given sZ, and the sequence order is the usual Z order. The 2-adic order cannot be used directly, but perhaps there is a bijection a:Z->Z whose value order is the 2-adic order? The value order of a:Z->Z is always the usual order of Z, not 2-adic. Wait a is a map from positions Z to values Z. The value order is usual. The order of positions by value is a bijection from values Z to positions Z, which has order type Z (positions usual). It cannot be the dense 2-adic order. So no. Let's look at the problem's wording "x1 <⋯ < xk". If the permutation is a bijection π:Z→Z, the terms of the permutation are π(n). A monotone AP in the permutation could be a sequence of terms π(x_1),...,π(x_k) where x_1<...Z has no monotone 3-AP. For d=1, π zigzags. Let's assume even indices are local minima, odd local maxima. Consider the subsequence of local maxima π(2m+1). For d=2, the even and odd subsequences zigzag. Thus the local maxima sequence (odd indices) has its own local maxima/minima at some parity of m. This creates a binary tree of extrema. Since values are integers, by an infinite Ramsey or pigeonhole, there is an infinite path through the tree where values are monotone? But the tree constraints force zigzag at each level, so no monotone triple at any level. However, an infinite path through a binary tree with alternating signs may exist (e.g., 2-adic). The values along a path corresponding to a fixed residue class modulo 2^r must zigzag; along a branch (a single arithmetic progression with difference 2^r?), the sequence over the branch is zigzag, so no monotone 3. This is consistent for all r. It resembles a labeling of the 2-adic tree by integers such that every branch is zigzag. Is that possible? We can label each node (integer position) by an integer; for each edge? The condition for step 1 says along every path (the integer line) signs alternate. For step 2, along each even/odd branch signs alternate. This is equivalent to: For each node n and each scale r, the sign from n to n+2^r depends only on the r-bit prefix? We can assign values recursively by choosing large increments at each level. It seems possible! For example, assign to each integer n a real number given by a convergent series of alternating terms based on its binary digits: q(n) = ∑_{r≥0} ε(n,2^r) 2^{-r}? This would be a 2-adic order (dense) and every branch zigzags. To make values integers, we can scale and choose coefficients to make all q(n) distinct integers? A countable set of such series can be mapped to integers via a strictly increasing enumeration? But the enumeration order would change the betweenness. However, if the series values are already distinct reals with the desired zigzag property, we can map them to integers by an order-preserving bijection only if the order type is Z. The order of these series values is likely dense, not Z. But we don't need to preserve the full order, only the betweenness for triples n,n+2^r,n+2^{r+1}. Maybe we can choose integer values recursively to realize the sign pattern without preserving all order relations. This is like solving a system of strict inequalities. The partial order may have a linear extension of type Z. We can try to construct one by assigning integer values in increasing order of a "rank" that is not the dense order. The partial order constraints are finite for each element? Each q(n) is constrained relative to q(n±d) for all d. Infinite. Maybe use a probabilistic method? Random permutation of Z? Probability of a monotone 3-AP? There are infinitely many APs; Borel-Cantelli? For a random permutation of Z (i.e., random assignment of ranks), the three values of an AP are equally likely to be in any of 6 orders; probability middle between endpoints is 2/6=1/3. There are infinitely many APs, so likely there is a monotone 3-AP. In fact, for any k, maybe random permutation has infinitely many monotone k-APs? Not helpful. Let's attempt a proof of no 3-free permutation of Z using the sign pattern and transitivity. Suppose σ(n,d)=sign(π(n+d)-π(n)) satisfies σ(n+d,d)=-σ(n,d). Can such a sign pattern be realized by a total order of type Z? The sign pattern may be forced to be σ(n,d)=(-1)^{v_2(d)}? Let's derive from transitivity of the total order for triples that are not APs? For any n,d,e, the order of n,n+d,n+e,n+d+e may impose constraints. The 3-free condition only applies to AP triples, but total order transitivity links signs. Maybe we can derive a contradiction using a parallelogram: Consider points n, n+d, n+2d, n+3d. The signs along step d alternate: σ(n,d)=+, σ(n+d,d)=-, σ(n+2d,d)=+ ... (WLOG). Thus the order of the four points in the AP must be: a0 < a2 < a1? Let's set a_i = n+id. Signs: a0a2; a2a2a4, and d=2 conditions, the order may be forced to a0a1, not a2 between. For non-adjacent a00 (0 min). AP 1,2,3: 2 extreme relative1,3. If 2<1, to avoid decreasing 1>2>3, need 3>2. Not force 2 vs0. AP 0,2,4: middle2. If 2<0, then 4>2; if 0<2<1, then 4? Could be. So 2 not forced between. Thus a 3-free order could have 2<0, leaving gap 0<1. Our immediate successor construction tried that and eventually dead-ended, but maybe not locally forced. To prove no discrete, need a global argument that the gap 0<1 eventually contains some element due to constraints involving other points. In our attempt, -3 was forced into gap 0<-1 when we chose 1<-1 adjacent. But if 1 not adjacent to -1, maybe no. Could there be a 3-free order of type Z where 0<1 adjacent and 2<0? Let's see if we can satisfy constraints without dead end by not making 1<-1 adjacent. Choose order ... 2<0<1< ... , and place -1 above 1 but not immediate; maybe put some elements between 1 and -1. What elements? 3? In AP 0,3,6? Not. We can leave room. A Z order requires every adjacent pair, but we can make 1's successor some x not -1, and -1 later above x. This may avoid forcing -3 into gap. Thus discrete order possible. Let's try to build a discrete order by choosing successor of 1 as 3, and place -1 above 3, etc. We tried 0<1<3<... and hit -3 issue when 4 was successor? Maybe if we don't make 3<4 adjacent, leave room. Let's attempt a systematic construction of a Z-chain that is "cofinal" in the 2-adic dense order: The dense 2-adic order has between 0 and1 infinitely many elements (powers of 2? in finite bit-rev, between 0 and1 are 4,2,6,...). To make a discrete order, we can insert a Z-chain that is cofinal in each interval but with immediate successors chosen from these elements. For example, choose the order sequence (by value) as: ... , 2, 4, 8, ..., 0, 1, 3, 5, 7, ...? Need 3-free. Maybe the discrete 3-free order can be the "2-adic order" with a chosen cofinal sequence in each cut. The 2-adic order is dense, but every countable dense order has a cofinal Z-chain? A Z-chain is a subset of type Z that is cofinal/coinitial in each interval? To extend the partial order, we need a linear extension, not a suborder of the dense order. We can take the dense order and "prune" it by choosing one element from each cut to be adjacent, while maintaining the partial order. This is like finding a discrete linear extension. Many dense partial orders have discrete extensions if they are not too thick. The 3-free partial order might be exactly the 2-adic order, which is dense and may not have discrete extensions because it's a total order; but the partial order (triple inequalities) may be a proper subset, allowing discrete extensions. Let's search for a theorem: "Every countable 3-free poset has a linear extension of type ω"? Not. Given time, we need decide final answer. Let's look at the user's question: "What is the largest k such that in any permutation of ℤ there must exist a monotone k-term arithmetic progression x1 <⋯ < xk?" It might be from a problem set expecting a concise answer. If the answer were 2, the main proof is existence of a 3-free permutation, which is quite advanced. If answer 3, proof of 3 unavoidable and 4 avoidable also advanced but perhaps known. Which is more likely to be asked as "largest k"? Maybe the answer is "3" because it's a sharp threshold and can be proved by a simple application of the Erdős–Szekeres theorem plus a construction. Let's try to find a simple proof of 3 unavoidable for permutations of Z (two-sided) using a known theorem: In any permutation of Z, there are two consecutive terms (in the permutation order) whose values differ by ±1? No, not true. But maybe in any permutation of Z, there exist three equally spaced terms that are monotone by a theorem of van der Waerden? Van der Waerden says any coloring of Z has monochromatic AP; not relevant. Could use the fact that the permutation order is a linear order of type Z. Consider the sequence of values at positions 0,1,2,... (one-sided ray). We proved no one-sided permutation of Z? Wait if the sequence is indexed by Z, the positive ray is a one-sided sequence of some subset of Z (not all). It may have no monotone 3-AP if 3-free. But a one-sided sequence of distinct reals with no monotone 3-AP can be arbitrarily long? We found finite; infinite one-sided of Z might exist (our greedy). If the full bi-infinite sequence is 3-free, its positive ray is a one-sided 3-free sequence of a subset of Z. That's possible. So no easy contradiction. Let's see if one-sided 3-free sequence of Z exists by continuing greedy. If it does, then a bi-infinite 3-free permutation may be easier: take two one-sided sequences on positive/negative positions with a central gap? APs crossing 0 need handle. But maybe we can take a one-sided 3-free sequence of Z and reflect it to negative positions: define a_n for n≥0 from greedy; for n<0, set a_n = -a_{-n-1}? This might be a permutation of Z and 3-free? Let's test. If a_n (n≥0) is a permutation of nonnegative? We need all integers. Suppose b_n (n∈N) is a bijection to Z? We can split: for n≥0 use positive and negative values; for n<0 use the rest. Cross APs with indices crossing 0 are tricky. Maybe a one-sided 3-free sequence of Z can be extended to two-sided by simply prepending the negative of the sequence in reverse? Let a_n (n≥0) be 3-free and a bijection to Z? A one-sided sequence cannot be a bijection to Z and 3-free? It could. Define q_n for n≥0 a bijection to Z with no monotone 3-AP. Then define q_{-n-1} = -q_n? That duplicates. Need disjoint. If a one-sided 3-free sequence of Z exists, then a permutation of Z indexed by N exists, so if the problem allows one-sided, answer 2. If problem requires two-sided, we can convert a one-sided sequence to two-sided by interleaving with a shifted copy? Let q_n (n∈N) be 3-free bijection to Z. Define p_n for n∈Z by p_{2n}=q_n, p_{2n+1}=q_{n+1}+C? Not permutation. Let's search memory of "3-free sequence of integers". I recall a simple construction: Let a_n = (-1)^n 2^{n} for n≥0? We saw even subsequence monotone. But maybe a_n = (-1)^{v_2(n)} 2^{n}? For n=1 -2, n=2 +4, n=3 -8, n=4 +16, n=5 -32... d=1 zigzag? -2<4>-8<16>-32 yes. d=2: even positions 4,16,64 monotone bad. Need decimations by 2 zigzag, so magnitudes for even positions must alternate sign. Use a_n = (-1)^{floor(n/2)} 2^n? n=1 +2? Let's define sign s(n)=(-1)^{floor(n/2)}: n=0 +,1 +,2 -,3 -,4 +,5 +,6 -,7 -. Values with magnitude 2^n: 1,2,-4,-8,16,32,-64,-128. d=1: 1,2,-4: 1<2>-4 non; 2,-4,-8: 2>-4>-8 decreasing (bad). Need signs alternate every 1 for d=1, so s(n)=(-1)^n. Then even subsequence all same sign if magnitude positive. To make even subsequence zigzag, magnitude for even n must itself have sign depending on floor(n/2). Thus sign s(n)=(-1)^n * (-1)^{floor(n/2)} = (-1)^{n+floor(n/2)}. This sign pattern for n=0:+,1:-,2:-,3:+,4:+,5:-,6:-,7:+ (Thue-Morse-like). Magnitudes could grow fast. Let's test with a_n = s(n) 2^n. n=0:1 1:-2 2:-4 3:8 4:16 5:-32 6:-64 7:128 d=1: 1,-2,-4 decreasing (1>-2>-4) bad. Need d=1 signs alternate: a0-2, a1>a2? -2>-4 false (decreasing). Magnitude dominates, so sign pattern not enough; need local extremum by magnitude: a1 should be max/min relative a0,a2. If a0=1,a1=-2,a2=-4, all negative trend. Choose magnitudes to make a1 extreme: e.g., a0=1,a1=-100,a2=-1? Then -100 min. This is our greedy. A recursive magnitude assignment can work: For each n, choose |a_n| large enough to make it a local extremum for all smaller d? At step n, constraints give an interval; choose a_n far outside all used values in the required direction if the interval is unbounded. If the interval is bounded, choose a value inside but leave room. This suggests an infinite construction is possible with a careful "always choose a new integer that leaves at least two unused integers in the allowed interval" strategy. We might be able to prove existence by induction: At step n, the allowed set is a (possibly finite) interval containing at least one unused integer. Choose a_n such that the allowed set for step n+1 contains at least two unused integers. Is that always possible? We need verify. This would give an infinite sequence (not necessarily surjective). To ensure surjective, we can later insert all integers using a diagonal priority, since there is always room. If we can establish existence of a 3-free permutation of Z (one-sided or two-sided), answer 2. Let's see if the problem asks for "largest k". If 3-free exists, k=2. But is k=2 always? In any permutation of Z, does there always exist a monotone 2-term AP? A 2-term AP is just two values aZ. A natural bijection splits N into evens/odds and Z into positives/negatives. Suppose a(2m) = m (nonnegative), a(2m+1) = -m-1 (negative). This is a bijection: evens get N, odds get negative integers. Sequence: 0,-1,1,-2,2,-3,3,-4,... Check d=1: 0,-1,1 non; -1,1,-2 non; 1,-2,2 non; d=2: even subsequence 0,1,2,3 monotone bad. If we rearrange evens to a 3-free one-sided sequence of N? But N one-sided no 3-free. However, evens domain N, range N; impossible if full 3-free. So parity split into one-sided ranges fails. Split N into two sets each mapped to Z (two-sided), not separated. For d=1, we need odd positions extreme between even neighbors. We can choose odd values very large in magnitude relative to neighboring even values. Let even positions be a permutation of, say, numbers with even 2-adic valuation, odd positions with odd? Maybe use magnitudes growing with position to dominate. Try a simple explicit formula: a_n = (-1)^n (n + 2^{v_2(n)} * n) = (-1)^n n (1+2^{v_2(n)}). For n=1: -2, n=2: +6, n=3:-12? v2(3)=0 => -6? Wait n*(1+1)=6, sign - => -6. n=4 v=2 =>4*5=20. Sequence -2,6,-6,20,-10,30,-14,56. d=1: -2<6>-6<20>-10<30>-14<56 good. d=2: even positions -2,-6,-10,-14 monotone decreasing (bad). Need even subsequence zigzag; factor for even n: n(1+2^{v2(n)}) with sign (-1)^n positive for even. It is increasing. If we alternate sign for even subsequence using (-1)^{floor(n/2)} as well: a_n = (-1)^n (-1)^{floor(n/2)} n(1+2^{v2(n)})? Sign pattern: n=0 +,1 -? floor1/2=0 => -, n=2 floor1 => +* -? (-1)^2=+, (-1)^1=- => -, n=3 (-1)^3=-, floor1 => - * - = +, n=4 +, floor2 + => +, n=5 -, floor2 + => -, n=6 +, floor3 - => -, n=7 -, floor3 - => +. Same as before. Magnitudes maybe enough to make d=1? n=0 1, n=1 -2, n=2 -4? -1,-2,-4 decreasing bad. Need a1 extreme relative a0,a2: -2 is between? 1>-2>-4 decreasing, bad. If magnitude at n=1 larger than n=2: e.g., a0=1,a1=-100,a2=-4, then -100 min. So odd magnitudes should dominate next even. We can set magnitude for odd n larger than neighboring evens. For d=2, even subsequence needs odd/even magnitudes among evens to dominate. This suggests a hierarchical magnitude: |a_n| should be large compared to |a_{n±d}| for d up to 2^r depending on n's 2-adic valuation. A formula with factor 2^{2^{v_2(n)}} might work. Let's try a_n = (-1)^n M(n) where M(n) = 2^{2^{v_2(n+1)}} * (n+1)? For n=0: +? v2(1)=0 => M=2^1*1=2? a0=2. n=1: v2(2)=1 => M=2^2*2=8, a1=-8. n=2: v2(3)=0 => M=2*3=6, a2=6. d=1: 2,-8,6: -8 min OK. n=3: v2(4)=2 => M=2^4*4=64, a3=-64. Triple -8,6,-64: 6 max OK. n=4: v2(5)=0 M=10, a4=10. Triple 6,-64,10: -64 min OK. d=2: even positions 2,6,10,? monotone increasing (bad). Need even subsequence zigzag: a0=2,a2=6,a4=10; middle a2=6 between -> monotone. We need sign/magnitude for even n to alternate: a2 should be extreme relative a0,a4. If a2 magnitude larger than both, e.g., a2=100, a0=2,a4=10, then a2 max. Thus for n=2 (v2(2)=1), magnitude should be larger than neighbors n=0,4. In general, n with high v2 (multiple of 2^r) should have large magnitude to be extreme for decimations of lower powers. This is promising: Let M(n) = 2^{2^{v_2(n)}} * (n+1) maybe, with sign (-1)^n? For n=0 v∞? handle. n=1 v0 small, n=2 v1 larger, n=4 v2 larger. Even subsequence: n=0 (small),2 (medium),4 (large) monotone increasing if signs same. Need sign alternate among even: use sign (-1)^{floor(n/2)} as well. Let sign s(n)=(-1)^n * (-1)^{floor(n/2)} = (-1)^{n+floor(n/2)}. Magnitudes M(n)=2^{2^{v_2(n)}} (n+1) (with v2(0)=∞ maybe huge). Compute n=0: v∞? set M0=1, s0=+ ->1. n=1: v0 M=2*2=4, s=- -> -4. n=2: v1 M=4*3=12, s? n+floor=2+1=3 -> - => -12. d=1: 1,-4,-12 decreasing (1>-4>-12) bad; a1 should be min, but -4 > -12. Need M1 larger than M2? Actually a1=-M1, a2=-M2; for a1 min need M1 > M2. But v2(1)=0 small, v2(2)=1 large, opposite. We want odd n to dominate neighboring evens for d=1, but even n=2 to dominate its even neighbors for d=2. This is hierarchical: at each scale, the element with certain residue should dominate. In finite bit-reversal, the element at index 1 (a1) has high value 8 in length16, dominating 0 and2? In length16: a0=0,a1=8,a2=4,a3=12. a1 high relative a0,a2; a2 high relative a0,a4? a2=4, a4=2, a0=0, a2 high; a3=12 high relative a1,a5? a3=12, a5=9, a1=8 high. The values are not simply based on v2(n); they follow bit-reversal: a_n's value is the reversed binary of n. The high bits correspond to low-order bits of n. For infinite, we can define a_n as a real/integer whose "magnitude hierarchy" is given by the 2-adic expansion. Perhaps a_n = the integer whose binary representation is the 2-adic expansion of n reversed, with a sign to make it integer. The value order (usual integer) would then realize the 2-adic order, which is dense. But if we only need 3-free, maybe we can take a_n = the rank of n in the 2-adic order within a finite prefix, and let rank tend to infinity? Not. Maybe there is no one-sided 3-free sequence of Z because the magnitude hierarchy would require an infinite decreasing chain of positive magnitudes? Our greedy used large positives and negatives, not one-sided. The finite bit-reversal values in [0,2^m) have a minimum and maximum; as m grows, to extend, the relative order of old values changes, so no nested sequence with bounded values. But with unbounded values, maybe possible. Let's try to prove existence of a one-sided 3-free sequence of Z by a known compactness argument. If we can present that in final answer, answer 2. The construction can be non-constructive: The set of finite 3-free permutations of finite sets has the property that every finite partial order can be extended; by the compactness theorem, there is a 3-free permutation of Z. Need ensure the range is Z (all integers). We can use first-order logic with constants for each integer and a unary function f:Z->Z? Let's formulate. We want a bijection q:Z->Z (positions? values?) such that for all x,d, q(x+d) is not between q(x),q(x+2d). This is an infinite first-order condition in the language with constant symbols for integers and a function symbol? Bijectivity is not first-order (surjectivity infinite). But we can use a language with a binary relation R (order of values) and require R is a linear order of type Z? Type Z not first-order. However, a permutation q:Z->Z is a bijection; we can encode it as a partial order? Maybe use compactness of the space of linear orders of type Z with the 3-free property. The class of linear orders of type Z is not first-order but is compact in the age sense? The set of finite orders (ages) that can be embedded in a Z-order? Any finite linear order can be embedded in Z. The 3-free property is a forbidden pattern? It forbids the order pattern where the middle of an AP is between endpoints. This is not just a finite order pattern on the three points? It involves the AP relation, which is a finite structure with domain and APs. The class of finite structures (finite subsets of Z with induced APs and a linear order) satisfying 3-free has arbitrarily large finite members (bit-reversal on intervals). By the compactness theorem for finite model theory (Erdős–Hajnal–Rado?), there is an infinite structure (Z with APs and a linear order) satisfying 3-free. But the linear order may be any countable linear order, not type Z. To get a permutation of Z, the linear order must be order-isomorphic to Z (discrete). The class of finite linear orders is the age of both Z and dense orders; compactness won't ensure discreteness. We can add a predicate for immediate successor? Discreteness is not a first-order condition but can be approximated: require that for each pair xZ directly as a sequence with values Z. The space of all bijections Z->Z is a Polish G_delta subset of Z^Z. The 3-free condition is closed? For each (n,d), the condition "q(n+d) is not between q(n),q(n+2d)" is closed in Z^Z? The set of triples where middle is between is open? Values discrete, so both are closed/open. The intersection over countably many is G_delta, possibly empty. Finite satisfiability doesn't guarantee nonempty in noncompact spaces. Maybe use a known result: The 3-free property is "compact" for permutations of Z because if every finite set of indices can be assigned distinct integers satisfying constraints on that set, then there is a global bijection by the compactness theorem for the space of integer orderings with one-point compactification. We can map Z to the compact interval by a homeomorphism? Not. Given uncertainty, perhaps we should look for the actual largest k in known problem. Let's search memory of exact phrase with "largest k". I recall a problem from the IMO Shortlist? "Let a permutation of the integers be a bijection π:Z→Z. A monotone arithmetic progression is a sequence π(a),π(a+d),...,π(a+(k-1)d) that is monotone. Determine the largest k such that every permutation contains a monotone k-term arithmetic progression." I think the answer was 3. There is a known solution: Every permutation of Z contains a monotone 3-term AP; construction avoiding 4 uses the sequence ... maybe from bit-reversal. This sounds like an olympiad/combinatorics problem. The use of Z (not N) is to make 3 unavoidable? Wait for N, we proved 3 unavoidable by well-ordering; for Z, maybe 3 still unavoidable due to a more subtle argument. The construction avoiding 4 for Z might use a two-sided version of bit-reversal blocks. The answer 3 is elegant. Let's try to prove 3 unavoidable for Z with a simple argument using adjacent values? Suppose π:Z->Z (sequence). If no monotone 3, then for every n,d, π(n+d) extreme. Consider the sequence of values at positions ...,-1,0,1,2,... . Since it's a permutation of Z, there is a value 0 at some position p. Translate positions so p=0. Then π(0)=0. For each d>0, the triple -d,0,d has middle 0, so π(-d) and π(d) must have the same sign (both positive or both negative). Thus the sets A={d>0: π(d)>0} and B={d>0: π(d)<0} partition positive d, and π(-d) has same sign as π(d). Now consider d=1. Suppose π(1)>0 (otherwise reflect). Then π(-1)>0. The triple 0,1,2 has middle π(1)>0, endpoints 0 and π(2). To avoid increasing 0<π(1)<π(2), we need π(2)<π(1). It could be positive less than π(1) or negative. The triple 1,2,3 has middle π(2). If π(2)<π(1), to avoid decreasing π(1)>π(2)>π(3), need π(3)>π(2). So the positive ray signs? We can derive that the sequence π(1),π(2),π(3),... zigzags starting with high? If π(1)>0 and π(2)<π(1). If π(2)>0, then π(3)>π(2) (could be >π(1) or between). If π(2)<0, then π(3)>π(2) (could positive). Not enough. But we also have symmetry: π(-1)>0. The triple -2,-1,0 has middle π(-1)>0, endpoint 0. To avoid decreasing? Values π(-2),π(-1),0 with π(-1)>0. For π(-1) max, need π(-2)<π(-1). Could be positive less or negative. The triple -3,-2,-1 etc. The negative ray zigzags. Maybe use the fact that π is a permutation: among the positive values, there is a smallest positive integer 1. Let its position be r. Similarly smallest negative -1. The sign symmetry around 0 implies if π(d)>0 then π(-d)>0; the positive values occur in symmetric pairs (d,-d) with possibly different magnitudes. The smallest positive value 1 occurs at some r and -r? Not necessarily same value; π(r)=1, π(-r)=s>0. Since 1 is the smallest positive, π(-r)≥1. If π(-r)>1, then consider the triple? The value 1 at r is a local? It may not be a local minimum in the sequence because adjacent values can be negative? If π(r)=1, its neighbors in position (r±1) could be negative or positive >1. For d=1, if r is local extremum? The sign pattern depends on parity. If 1 is a local maximum? It cannot be > neighbors if neighbors positive? It could be if neighbors negative. If 1 is a local minimum, neighbors positive >1 or negative <1. This might lead to contradiction using the global smallest positive. Let's attempt: Let p be position of 1. Since 1 is the smallest positive, any negative value is <1. For the 3-free condition at position p with d=1: the triple p-1,p,p+1 has middle 1. It must be extreme. It cannot be max relative to endpoints if either endpoint is negative? Actually 1 > negative, but if the other endpoint is positive >1, then endpoints one negative one positive, 1 between? Example -1,1,2: 1 between -1 and2 -> monotone increasing? -1<1<2 bad. Thus for 1 to be extreme, both endpoints must be negative (1 max) OR both endpoints positive and 1 min (but endpoints positive >1). If both endpoints positive >1, 1 min OK. If one endpoint negative and one positive >1, bad. If both endpoints negative, 1 max OK. Since 1 is smallest positive, endpoints cannot be positive equal; if positive, >1. Thus the two neighbors of 1 are either both negative or both positive (>1). Similarly for -1 (smallest magnitude negative? -1 is greatest negative). Let q be position of -1. The triple around q: endpoints must be either both positive ( -1 min) or both negative < -1. Can we derive a contradiction from the positions of 1 and -1? They are distinct positions. Consider the AP in positions with midpoint between them? If p and q have same parity, m=(p+q)/2 integer. The values at p,q are 1,-1. The middle value at m is some integer. For the triple (m? actually p,q endpoints, m middle) to be 3-free, the middle value must be extreme relative 1 and -1. Since -1<1, the middle value must be either < -1 or >1 (cannot be 0 or between? 0 is between -1 and1, would make monotone? If values -1,0,1 middle 0 between bad. So middle not in [-1,1]). Thus π(m)≤-2 or ≥2. If p and q have different parity, no midpoint. But consider the APs with difference 1 around p and q; maybe. The values 1 and -1 are special. Since the permutation is of Z, there is also 0 at position 0 (we translated). The triple (-1,0,1) in positions required π(-1) and π(1) same sign. Thus 1 and -1 positions? Not. Maybe use the positions of the three values -1,0,1. If their positions form an AP, then they are monotone (values -1<0<1 or reverse) if positions monotone? The 3-free condition is on position APs, not value APs. If positions of -1,0,1 are an AP, then the values are monotone? Suppose positions p,q,r with q middle in positions; values at p,q,r are some permutation of -1,0,1. The middle position's value is extreme? For values -1,0,1, the middle value 0 is between the other two, so if the position middle has value 0, bad; if position middle has value -1 or1, OK. The values -1,0,1 as a set can be placed to avoid. Not contradiction. This is too hard. Let's consider the possibility that the answer is actually infinite? In any permutation of Z, must there exist monotone k-AP for all k? No, because there are permutations with bounded longest monotone AP? Maybe for Z, by a theorem, there are permutations with no monotone 4, so finite. If 3-free exists, answer 2. If not, 3. Could it be 4? Every permutation has monotone 3 and 4, but can avoid 5? The question asks largest forced; maybe answer 3 or 2. Unlikely 4 because finite 3-free permutations show no finite forcing of 3 in one-sided? But infinite may force 3; forcing 4 would be stronger. Can we avoid 4 easily? Likely yes, so not 4. Let's search memory of "monotone 3-term arithmetic progression permutation integers". I recall a result: "Every permutation of the positive integers contains an increasing arithmetic progression of length 3"? Wait if values are positive and positions N, increasing only? The decreasing permutation of positive integers doesn't exist for N (can't list all positives decreasing). But an increasing 3-AP might be unavoidable. For Z, decreasing all values avoids increasing APs, so they must allow both. The term "monotone" often means either increasing or decreasing. There is a theorem: Every permutation of the integers contains a monotone 3-term arithmetic progression. This is Theorem 1 in a paper "Monotone arithmetic progressions in permutations" by? And they construct a permutation with no monotone 4-term AP. The largest k is 3. I think this is likely. Let's try to outline a proof for 3 unavoidable that we can present succinctly, even if not fully rigorous? We need final answer credible. We can state known result? The user likely expects answer with proof. We need provide enough detail. Maybe there is a simple proof of 3 unavoidable using the following lemma: Lemma: In any permutation of Z, there exist two consecutive terms (in the permutation order) whose indices differ by 1? Wait permutation order is positions; consecutive terms in the permutation order are positions? If the permutation is a bijection π:Z->Z, the terms are ordered by positions. Consecutive terms are π(n),π(n+1). Their indices differ by 1. Not helpful. Another lemma: For any permutation π of Z, there exists d such that the sequence π(n),π(n+d),π(n+2d) is monotone. This is exactly 3. Could prove by contradiction using the sign pattern and the fact that the sign pattern defines a 2-coloring of the complete graph? Maybe use Ramsey's theorem on the signs σ(n,d). For d=1, signs alternate. For each n, define the sign of the difference to the next local extremum. There are infinitely many local maxima/minima. By Ramsey, there is a 3-AP of local maxima with monotone values? The local maxima are at positions of one parity; their values form a sequence with no monotone triple for step 2? Actually the subsequence of local maxima (say odd positions) has no monotone triple with indices differing by 2? For d=2, the odd subsequence zigzags, so no

What is the largest 𝑘 such that in any permutation of ℤ the...